^OF-CALIFO%, 


THE  UNIVERSITY  OF  CHICAGO 
MATHEMATICAL  SERIES 

ELIAKIM  HASTINGS  MOORE 

GENERAL  EDITOR 


SCHOOL  OF  EDUCATION 

TEXTS  AND   MANUALS 

GEORGE  WILLIAM  MYERS 

EDITOR 


SECOND-YEAR  MATHEMATICS 
FOR    SECONDARY    SCHOOLS 


THE  UNIVERSITY  OF  CHICAGO  PRESS 
CHICAGO,  ILLINOIS 


Hgents 
THE  BAKER  &. TAYLOR  COMPANY 

NEW    YORK 


CAMBRIDGE  UNIVERSITY  PRESS 

LONDON    AND    EDINBURGH 


Second -Year  Mathematics 

For  Secondary  Schools 


By 

GEORGE  WILLIAM  MYERS 

Professor  of  the  Teaching  of  Mathematics  and  Astronomy,  College  of 
Education  of  the  University  of  Chicago 

and 

WILLIAM  R.  WICKES  ERNEST  A.   WREIDT 

ERNST  R.   BRESLICH  ARNOLD  DRESDEN 

Assisted  by 
ERNEST  L.   CALDWELL  and  ROBERT  M.   MATHEWS 

Instructors  in  Mathematics  in  the  University  High  School 
of  the  University  of  Chicago 

2  /  2  8  3~ 


SCHOOL  OF  EDUCATION   MANUALS 
SECONDARY  TEXTS 


THE  UNIVERSITY   OF  CHICAGO  PRESS 
CHICAGO,  ILLINOIS 


All  Rights  Reserved 


Published  June  1910 
Second  Impression  October  1910 


Composed  and  Printed  By 

The  University  of  Chicago  Press 

Chicago,  Illinois.  U.S.A. 


PREFACE 

This  book  carries  forward  through  the  second  high-school 
year  the  combined  type  of  material  and  the  plan  of  treatment 
of  First-Year  Mathematics.  The  two  texts  together  cover  the 
essentials  of  what  is  commonly  required  of  all  pupils  in  the 
first  two  years  of  secondary  schools  in  this  country,  and  include, 
in  addition,  the  elementary  notions  of  plane  trigonometry 
through  the  solution  of  right  triangles,  as  well  as  an  introduc- 
tion to  some  topics  of  formal  algebra  not  usually  treated  iifc. 
secondary  texts.  Each  book  constitutes  a  well-balanced  and 
not  over-heavy  year  of  work.  This  material  so  arranged  at 
the  same  time  opens  to  the  pupil  a  broader,  richer,  a  more 
useful  and  therefore  a  more  alluring  field  of  ideas  than  do  the 
two  subjects  of  algebra  and  plane  geometry  treated  separately. 

It  is  felt  that  the  material  and  treatment  of  these  books 
lays  for  the  beginner  a  more  stable  foundation  for  future  work 
than  does  the  usual  order  of  a  year  of  formal  algebra,  followed 
by  a  year  of  formal  demonstrative  geometry,  or  of  these  sub- 
jects in  the  reverse  order.  This  judgment,  founded  on  our 
own  experience,  is  confirmed  by  a  recent  extended  inspection 
of  schools  abroad.  In  England  as  well  as  in  Germany  and 
France,  the  best  secondary  schools  were  seen  to  be  using  the 
methods  of  combined  mathematics  almost  exclusively,  with 
seeming  advantage  to  their  pupils. 

Second-Year  Mathematics  lays  chief  emphasis  on  geometry, 
as  did  the  First-Year  Mathematics  on  algebra.  To  take  up 
the  work  of  these  texts  then  requires  no  abrupt  departure  from 
the  order  of  subjects  now  prevailing  in  secondary  curricula. 
The  plan  of  the  books  enables  the  work  of  the  first  high-school 
year  to  connect  smoothly  and  strongly  with  eighth-grade  work 


viii  .     Preface 

through  both  mensuration  and  general  number,  rather  than 
with  one  of  these  subjects  in  the  first  year,  and  the  other 
subject  in  the  second  year. 

First-Year  Mathematics  thus  becomes  a  continuation  and 
an  outgrowth  of  these  two  arithmetical  topics;  and,  without 
losing  hold  on  geometrical  notions  already  begun,  it  develops 
the  customary  topics  of  first-year  algebra  well  into  quadratics. 
Toward  the  close  of  the  first  year  geometrical  ideas  are  revived 
and  considerable  preliminary  geometrical  work  is  done. 

The  second-year  book  then  begins  with  some  constructive 
and  inductive  geometry,  and  passes  rapidly  to  demonstrative 
geometry,  employing  for  a  time  the  half-experimental  method 
of  superposition.  By  the  employment  of  algebraic  notation 
and  by  the  continued  application  of  the  equation  to  geo- 
metrical matters,  the  hold  on  algebra  is  kept  firm  until  the 
opportunity  arises  to  develop  with  profit  other  algebraic  topics, 
such  as  a  completion  of  methods  of  solution  of  the  quadratic 
equation,  a  discussion  of  the  roots,  and  the  use  of  inequalities 
in  the  solution  of  indeterminate  equations.  In  the  second 
year,  therefore,  the  algebraic  ground  already  gained  is  not 
only  held,  but  is  extended  at  least  as  far  as  is  customary  with 
the  algebra  before  the  third  year.  The  quadratic  equation  is 
used  from  time  to  time  throughout  the  second  year.  All  of 
plane  geometry  is  taught  with  sufficient  fulness.  Thus,  while 
the  first  book  may  be  styled  algebra  with  associated  arith- 
metic and  geometry,  the  second  may  be  styled  geometry  with 
associated  algebra  and  trigonometry. 

The  plan  is  continually  to  command  a  retrospect  and 
prospect  from  the  field  of  view  of  algebra  during  the  first  year 
and  a  retrospect  and  prospect  from  the  field  of  view  of  geome- 
try during  the  second  year.  Nowhere  is  the  horizon  to  be 
needlessly  restricted,  nowhere  are  familiar  fields  to  be  too 
soon  lost  to  view,  and  nowhere  is  an  exhilarating  outlook  and 
forecast  to  be  denied.  Oftentimes  a  high-school  pupil  fails 


x 

rightly  to  esteem  a  school  subject  because  he  cannot  discern 
its  bearings  on  what  has  preceded  and  on  what  is  to  follow,  or 
even  on  the  demands  of  the  time.  These  books  attempt  to 
bring  to  the  early  mathematical  subjects  the  benefits  of  some 
knowledge  of  these  bearings. 

The  early  and  rather  full  use  of  the  principles  of  congruency 
(chap,  i),  of  proportionality  (chaps,  ii  and  iv),  of  measurement 
(chap,  iii),  and  of  inequality  (chap,  v),  is  worthy  of  remark. 
This  sequence  of  principles  will  be  appreciated  by  many  as 
giving  to  the  pupil  in  the  order  of  difficulty  a  good  grasp  of  the 
four  chief  agencies  of  geometrical  reasoning.  More  than 
90  per  cent,  of  the  theorems  of  school  geometry  are  proved  by 
them.  The  early  part  of  four  of  the  first  five  chapters  thus 
adds  a  new  geometrical  tool  to  the  equipment  of  the  pupil. 
Furthermore,  these  new  tools  are  added  in  close  enough 
proximity  to  enable  each  one  to  be  seen  in  the  light  of  com- 
parison and  contrast  with  others.  It  is  pedagogically  a  waste 
if  not  an  impossibility  to  try  to  teach  well  one  and  only  one 
idea  or  method  at  a  time.  These  principles  are  not  taught 
explicitly  as  geometric  methods,  they  are  taught  by  making 
each  one  in  turn  the  predominant  method  in  demonstrations 
long  enough  to  give  the  pupil  a  clear  idea  of  its  meaning,  and 
a  practical  control  of  its  use. 

The  authors  are  convinced  that  a  topical  treatment  of  the 
theory  of  limits  does  not  belong  to  the  early  years  of  a  high- 
school  course,  and  in  this  they  are  in  harmony  with  the  writers 
of  some  of  the  best  among  recent  French  and  German  second- 
ary texts.  What  has  been  attempted  here  is  to  pave  the  way 
toward  a  later  thorough  understanding  of  this  important  sub- 
ject, rather  than  to  induce  the  pupil  to  think  he  has  mastered 
the  subject  by  the  memorizing  of  a  theorem.  The  question  of 
the  existence  of  incommensurable  lines  and  numbers  is  raised; 
examples  of  these  are  given;  their  approximations  by  means  to 
continued  fractions  as  well  as  by  decimal  fractions  are  studied; 


x  Preface 

whether  ratio-theorems  can  be  proved  to  hold  true  when  such 
magnitudes  are  to  be  compared,  is  discussed;  the  error  made 
in  using  a  rational  number  as  an  approximation  to  the  common 
measure  of  two  incommensurable  numbers  is  evaluated  and 
shown  to  grow  less  and  less  as  the  process  of  approximation 
is  continued,  calling  forth  intentionally  the  conviction  that 
there  is  a  definite  number  toward  which  these  approximations 
tend.  Finally,  in  the  treatment  of  the  ratio  of  the  diameter 
of  a  circle  to  the  circumference,  the  notion  of  the  limit  of  a 
sequence  is  more  fully  developed,  and  it  is  hoped  that  as  a 
result  of  the  preliminary  work  done  throughout  the  book, 
the  pupil  may  now  gain  some  fairly  clear  notion  of  that 
concept. 

This  simple  and  sane  treatment  of  incommensurables;  the 
concrete  or  experimental  approach  to  new  topics  and  to  the 
subject  as  a  whole;  the  careful  treatment  of  similarity  and  the 
extension  of  it  into  the  trigonometry  of  right  triangles;  the 
reduced  number  of  major  propositions  carefully  though  not 
too  fully  treated,  and  the  basing  of  other  theorems  upon  them; 
the  uses  of  the  graph,  and  the  chapter  on  geometric  algebra, 
are  some  features  of  this  text  that  will  be  appreciated  by 
students  of  the  mathematical  needs  of  secondary  schools. 

The  critical  reader  will  find  in  this  text  many  places  where 
rigor  is  lacking.  No  one  can  be  more  conscious  of  the  exist- 
ences of  these  places  than  are  the  authors.  But  this  book  is 
for  boys  and  girls,  many  of  whom  nowadays  reach  the  second 
high-school  year  at  fourteen.  They  are  yet  only  children,  and 
the  degree  of  rigor  only  has  been  attempted  that  is  appreciable 
and  attainable  by  the  larger  number  of  them.  Scientific  rigor 
is  the  limit  toward  which  the  high-school  pupil  approaches 
step  by  step,  but  which  he  cannot  reach  by  means  of  ideals 
so  far  beyond  his  immature  standards  that  he  cannot  even 
appreciate  that  they  are  ideals.  Criticisms  on  the  point  of 
lack  of  rigor  can  then  be  looked  upon  only  as  differences  of 


Preface  xi 

judgment  as  to  what  is  appreciable  and  attainable  rigor  by 
fourteen-year-old  boys  and  girls.     . 

This  text  is  published  primarily  for  use  in  the  classes  of 
the  University  High  School.  This  school,  however,  possesses 
no  features  that  call  for  class  material  in  any  way  different 
from  that  needed  by  all  good  public  high  schools.  It  would 
seem  then  that  the  book  must  be  of  interest  to  all  teachers  of 
secondary  mathematics  who  are  seeking  to  improve  their 
work.  The  authors  wish  to  emphasize  that  these  books  are 
an  attempt  at  finding  a  form  of  secondary  texts  in  accord  with 
the  pedagogical  and  mathematical  advance  of  recent  years. 
The  application  of  this  advance  to  secondary  problems  is  as 
yet  in  an  experimental  stage,  therefore  these  texts  should  not 
be  looked  upon  as  final  in  the  field,  but  rather  as  a  reflection 
of  the  present  evolutionary  stage  of  secondary  mathematics, 
to  be  perfected  as  the  study  of  the  problems  of  secondary 
mathematics  advances.  Criticisms  and  suggestions  looking  to 
the  better  carrying  out  of  the  plan  are  therefore  invited.  The 
problem  of  getting  better  texts  in  mathematics  for  high  schools 
is  quite  as  much  a  practical  as  a  theoretical  one,  and  the 
co-operation  of  teachers  is  earnestly  sought. 

In  conclusion,  the'  authors'  acknowledgments  are  due  to 
those  in  administrative  relations  with  the  University  High 
School,  especially  Charles  H.  Judd,  director  of  the  School  of 
Education,  and  Franklin  W.  Johnson,  principal  of  the  Univer- 
sity High  School,  and  also  to  Eliakim  Hastings  Moore,  pro- 
fessor and  head  of  the  Department  of  Mathematics  of  the 
University  of  Chicago;  for  their  sympathy  and  encouragement 
in  the  experimental  work  out  of  which  this  book  has  grown, 
and  for  their  very  substantial  help  in  bringing  this  publication 
to  the  light  of  day. 

THE  AUTHORS 

June,  1910 


TABLE  OF  CONTENTS 

. 

CHAPTER  PAGE 

I.     CONGRUENCY   OF   RECTILINEAR  FIGURES   AND   CIRCLES  I 

Circles 46 

II.  RATIO,  PROPORTION,  SIMILAR  TRIANGLES     ....  63 

Ratio  of  Numbers  and  Segments 63 

Ratio  of  Other  Magnitudes 64 

Commensurable  and  Incommensurable  Segments 

and  Magnitudes 66 

Proportion 69 

Problems  of  Construction 81 

Method  of  Analysis 86 

Algebraic  Exercises 87 

Similar  Triangles 89 

Similar  Polygons 99 

Similar  Right  Triangles,  Trigonometry  ....  102 

Problems  on  Right  Angles 108 

Relations  of  Trigonometric  Ratios 116 

Problems  and  Exercises 118 

III.  THE  MEASUREMENT  OF  ANGLES  BY  ARCS  OF  THE  CIRCLE  124 

Inscribed  Angles 130 

The  Method  of  Common  Factors 142 

The  Method  of  Successive  Division 143 

IV.  SIMILARITY  AND  PROPORTIONALITY  IN  CIRCLES  .     .     .  146 

To  Draw  Common  Tangents.     Second  Method     .  157 

Solution  of  Quadratic  Equations  by  the  Formula  .  161 

The  General  Quadratic  Formula 163 

Problems  in  Physics  Leading  to  Complete  Quad- 
ratic Equations 166 

V.  INEQUALITIES  IN  TRIANGLES  AND  CIRCLES      .     .     .  168 

xiii 


xiv  Table  of  Contents 

CHAPTER  PAGE 

VI.  AREAS  OF  POLYGONS 207 

The  Area  of  the  Triangle 207 

Areas  of  Polygons 229 

Proof  by  the  Method  of  Analysis 234 

Quadratic  Equations  in  Two  Unknowns      .      .      .  237 

Quadratic  Equations  Solved  by  the  Graph        .      .  239 

VII.  REGULAR    POLYGONS    INSCRIBED    IN,    AND    CIRCUM- 
SCRIBED ABOUT,  A  CIRCLE 246 

The  Side  of  a  Regular  Inscribed  Decagon  .      .      .  253 

The  Side  of  a  Regular  Inscribed  Pentagon  .      .      .  253 
Circles  Circumscribed  about,  and  Inscribed  in,  a 

Regular  Polygon 257 

To  Find  the  Circumference  of  a  Circle  ....  258 

Area  of  the  Circle        267 

VIII.   PROBLEMS  AND  EXERCISES  IN  GRAPHIC  AND  GEOMETRIC 

ALGEBRA 271 


CHAPTER  1 

CONGRUENCY   OF   RECTILINEAR   FIGURES    AND    CIRCLES 
2-  /  2.  £>O" 

1.  A  theorem  is  a  statement,  the  truth  of  which  is  to  be 
proved. 

EXAMPLE:  If  two  angles  are  equal,  their  supplements  are  equal. 

2.  A  problem  is  a  statement  of  a  construction  to  be  made, 
and  to  be  proved  correct. 

EXAMPLE:  To  draw  a  line  perpendicular  to  a  given  line  at  a  given 
point  upon  it. 

3.  A  proposition  is  either  a  theorem  or  a  problem. 

A  theorem  consists  of  two  parts,  the  hypothesis  and  the 
conclusion. 

The  hypothesis  is  what  is  assumed  to  be  true;  and  the 
conclusion  is  what  is  to  be  proved  to  be  a  consequence  of  the 
hypothesis. 

In  the  theorem  given  above,  "two  angles  are  equal"  is  the  hypothesis; 
and  "their  supplements  are  equal"  is  the  conclusion. 

4.  A  construction  is  an  exercise  to  draw  some  geometrical 
figure  without  reference  to  a  proof. 

CONSTRUCTION  I 

5.  To  draw  a  circle  of  given  center 
and  radius. 

Let  O  (Fig.  i)  be  the  given  center 
and  let  the  line,  r,  be  the  given  radius. 

Spread  the  compasses  until  the 
distance  between  the  tips  is  r. 

Place  the  pin-tip  on  O  and  draw 
the  pencil-tip  entirely  around  from  A, 
through  B,  C,  D,  to  A. 


Second-  Year  Mathematics 


FIG.  2 


6.  A  plane  figure  bounded  by  a  curve,  every  point  of  which 
is  equally  distant  from  a  point  within  the  curve,  is  a  circle. 

The  circle  is  sometimes  defined  as  the  curve  that  bounds 
the  figure  (see  FYM*  p.  200). 

7.  The  point  within  the  curve,  as  O  (Fig.  i),  is  the  center. 

EXERCISES  IN  CONSTRUCTION 

i.  Draw  a  circle  with  any  convenient  radius.  Without 
changing  the  distance  between  the  compass- 
points  step  around  the  circle,  and  mark  the 
steps.  Connect  the  alternate  marks  and 
form  a  trilobe  as  shown  in  Fig.  2. 

2.  Draw  a  six-lobed  figure  as  shown  in 
Fig.  3- 

3.  Mark  a  point  on  paper  and  draw  a 
circle    about   this  point    as   center   with    a 
radius  of  i  inch;   of  £  inch;   of  £  inch;    of 
ij  inches. 

4.  Mark  a  point  on  the  blackboard,  and 
with  crayon  and  string  draw  a  circle  about 
this  point  as  center  with  a  radius  of  i  foot; 
of  9  inches;   of  6  inches;  of  15  inches. 

8.  The  circles  of  either  of  the  Exercises  3  and  4  are  called 
concentric  circles  because  they  have  a  common  center. 

CONSTRUCTION  II 

9.  Given  the  three  sides  of  a  triangle,  to  construct  the  triangle. 
Let  a,  b,  and  c  (Fig.  4),  be  the  three  sides  of  the  triangle. 
To  draw  the  triangle. 

Draw  a  straight  line,  as  A  X,  longer  than  the  side  to  be 
first  laid  off,  say  c. 

*  The  abbreviation,  FYM,  stands  for  First-Year  Mathematics. 


FIG.  3 


Congruency  of  Rectilinear  Figures  and  Circles  3 

With  A  as  center  and  with  c  as  radius,  mark  an  arc  across 
AX,asatB.  ThenAB=c. 

With  A  as  center  and  with  radius  a,  mark  an  arc,  as  i, 
so  that  some  of  its  points  will  look  farther  from  B  than  the 
length  b,  while  other  points  of  arc  i  will  look  nearer  to  B 
than  the  length  of  b. 


IB 
FIG.  4 

With  B  as  center  and  with  b  as  radius  mark  an  arc,  as  2, 
across  arc  i. 

Connect  the  crossing-point,  C,  of  arcs  i  and  2  with  A  and 
with  B  by  straight  lines.  Then  A  B  C  is  the  required  tri- 
angle. 

10.  A  portion  of  a  plane  completely  bounded  by  three 
straight  lines,  is  a  triangle. 

EXERCISES  IN  CONSTRUCTION 

A  triangle  having  two  equal  sides  is  an  isosceles  triangle; 
the  third  side  is  the  base. 

1.  Draw  a  triangle  having  each  of  its  sides  i  foot  long. 

A  triangle  having  all  sides  equal  is  an  equilateral  triangle. 

2.  Which  side  is  the  base  of  the  triangle  in  Exercise  i  ? 

A  triangle,  no  two  of  whose  sides  are  equal,  is  a  scalene 
triangle. 

Any  side  of  a  scalene  or  of  an  equilateral  triangle  may  be  regarded 
as  the  base. 


Second-Year  Mathematics 

3.  Construct  an  equilateral  triangle  on  a  side  15  inches  long. 

4.  Is  an  equilateral  triangle  isosceles?     Give  reason  for 


answer. 


5.  Draw  a  triangle  having  sides  of  £  inch,  f  inch,  and 
i  inch,  beginning  by  laying  off  the  shortest  side  first. 


60' 

FIG.  5 

6.  With  crayon  and  string  draw  on  the  blackboard  a  tri- 
angle having  sides  6  inches,  8  inches,  and  10  inches. 

7.  Describe  how  with  a  66-foot  chain  stakes  may  be  set  in 
the  ground  at  the  corners  of  a  triangle  of  sides  30',  50',  and 
60'*  (see  Fig.  5). 

8.  Draw  a  triangle  having  sides  12",  12",  and  16". 

The  definite  portion  of  a  line  included  between  two  points 
is  a  line -segment.  For  example,  the  sides  of  a  triangle  are 
line-segments. 

CONSTRUCTION  III 

ii.  Given  the  base  and  one  of  the  equal  sides  of  an  isos- 
celes triangle,  to  construct  the  triangle. 

Take  a  line-segment  i  inch  long  for  the  base  and  another 
line-segment  £  inch  long  for  the  given  side,  and  construct  the 
triangle  with  compasses  on  paper,  or  instead  of  the  "inch"  use 
"foot,"  and  construct  the  triangle  with  crayon  and  string  on 
the  blackboard. 

*  The  single  prime  ('),  when  written  beside  an  arithmetical  number, 
means  foot,  or  ]eet;  the  double  prime  (")  means  inch  or  inches. 


Congruency  of  Rectilinear  Figures  and  Circles  5 

The  word  "line"  here  and  from  now  on  means  straight 
line,  unless  some  other  meaning  is  expressly  stated. 

A  line  is  supposed  to  be  indefinitely  extended  in  both 
directions. 

EXERCISES 

1.  The  span  of  a  roof  is  24'  and  one  of  the  two  equal  rafters 
is  15'.    Make  a  drawing  to  a  convenient  scale  of  the  end 
rafters. 

2.  One  of  two  equal  lines  is  denoted  by  #+12,  and  the 
other  by  i$x.     Find  x,  and  the  length  of  the  equal  lines. 

3.  Find  the  unknown,  x,  and  the  length  of  the  equal  sides 
of  an  isosceles  triangle  if  one  of  the  equal  sides  is  x3—  $x  and 
the  other  is  5^  +  9. 

4.  How  many  isosceles  triangles  will  answer  the  conditions 
of  Exercise  3  ?     Sketch  the  triangles. 

5.  Find  x,  and  the  two  equal  sides  of  an  isosceles  triangle, 
the  equal  sides  being  denoted  by  the  following  pairs  of  num- 
bers: 

(1)  x2—  x      and  $x—  8  (6)  xa+$x         and  4(15—  #) 

(2)  x2—  x      and  x+i$  (7)  x2—  7  and  3(27—3;) 

(3)  xa  and  7—  6x  (8)  ^x^+^x       and  12(1—  x) 

(4)  x(x+4)  and  3(2^  +  5)       (9)  $x(x+8)      and  2(20  +  2^) 

(5)  #(#—3)  and  i2(x—  3)  (10)  3^(10^—3)  and  20(2*—  i). 

CONSTRUCTION  IV 

12.  Given  a  side,  to  construct  an  equilateral  triangle, 
The  solution  is  left  to  the  student. 

EXERCISES 
i.  The  sides  of  an  equilateral  triangle  may  be  denoted  by 


2(5—x)>  3y~2>  and  2y-     Find  x>  y>  and  tne  length  of  a  side 
of  the  triangle. 


Second-Year  Mathematics 


2.  Find  x,  y,  and  the  length  of  a  side  of  an  equilateral 
triangle,  the  sides  of  one  being  designated  by  each  of  the 
following  sets  of  numbers: 

(1)  i2X,  (2+y),  and  3(6^-5)      (3)  4*,  2(^+3),  and  5(4-?) 

(2)  3*,  29-5^,  and  19  +  5? 

(5)  nx, 


—  A.y 
' 


1C 

V 


(4)  x,  y~3,  and 
,  and  —11(2^+9). 

CONSTRUCTION  V 

13.  Given  two  sides  and  twice  the  third  side  of  a  triangle, 
to  construct  the  triangle. 

_  a  _  Let  a,  b,  and  2C  be  the  given 

line-segments. 

To  construct  a  triangle  having 
a,  b,  and  c  as  sides. 

Let  us  first  find  one-half  of 
the  line  2c,  i.  e.,  let  us  bisect  2C. 
Make  AB  =  2c. 

With  A  as  a  center  and  with 
a  radius  longer  than  half  of 
A  B  draw  arcs  i  and  2  (Fig.  6). 
With  B  as  a  center  and 
with  the  same  radius  draw  the 
arcs  3  and  4  across  the  arcs  i  and  2,  respectively.  Call  the 
crossing-points  C  and  D. 

Join  C  and  D  and  call  E 
the  point  where  the  lines  C  D 
and  A  B  cross. 

A  E,  or  E  B,  is  c,  for  either 
is  one-half  of  2c. 

Now  construct  a  triangle  with 
a,  6,  and  A  E,  or  E  B,  as  sides 
as  in  Construction  II. 


FIG.  6 


,'tf 


l 


A 


>* 


£ 
FIG.  7 


Congruency  of  Rectilinear  Figures  and  Circles  7 

Is  it  necessary  that  arcs  2  and  4  of  Fig.  6  have  the  same 
radii  as  arcs  i  and  3  ?  (See  Fig.  7.) 

Is  it  necessary  that  the  intersections  lie  on  opposite  sides 
of  the  line  to  be  bisected  ? 

If  the  radii  of  the  arcs  i  and  2  of  Fig.  7  are  equal,  what  is 
the  only  essential  for  the  arcs  3  and  4  ? 

OBSTRUCTION  VI 

14.  To  construct  on  a  line  an  angle  equal  to  a  given  angle, 
and  having  its  vertex  at  a  given  point  on  the  line. 
See  Fig.  8  and  FYM,  p.  148. 


15.  The  figure  formed  by  extending  two  lines  from  a  point 
is  an  angle.     Thus  O  A  and  O  B  (Fig.  9)  extending  from  the 


point  O,  form  the  angle  A  O  B,  and  O'A'  and  O'B',  extending 
from  the  point  O',  form  the  angle  A'O'B'. 


8 


Second-  Year  Mathematics 


The  lines  that  form  the  angle  are  the  sides,  or  arms,  and 
the  point  from  which  the  lines  extend  is  the  vertex  of  the  angle. 

The  magnitude,  or  size,  of  an  angle  is  the  amount  of  turn- 
ing necessary  to  rotate  a  line  around  the  vertex  from  one  arm 
to  the  other. 


sir 


A  straight  angle  is  an  angle  whose  arms  lie  in  the  same 
straight  line  on  opposite  sides  of  the  vertex  (Fig  rr>)  '  The 
two  straight  angles  about  a  point  make  a  perigon.  A  perigon 
corresponds  to  a  complete  turn  of  a  rotating  line  (Fig.  n). 


A  right  angle  is  one-half  of  a  straight  angle  (see  Fig.  12,    I 


An  acute  angle  is  an  angle  that  is  less  than  a  right  angle 
(see  Fig.  12,  angle  a). 


An  obtuse  angle  is  an  angle  that  is  greater  than  a  right 
angle  and  less  than  a  straight  angle  (see  Fig.  12,  angle  c).  __ 


CONSTRUCTION  VII 

1 6.  Given  two  sides  and  the  included  angle  of  a  triangle,  to 
construct  the  triangle. 

Given  the  sides  a  and  b,  and  the  angle  C  (Fig.  13). 


Congruency  of  Rectilinear  Figures  and  Circles  9 

To  construct  the  triangle. 

Make  C  A  =  6,  and  at  one  end,  as  at  C,  construct  an  angle 
equal  to  the  given  angle  C  (see  Construction  VI). 


Make  C  B  equal  to  a. 

Connect  B  and  A. 

A  B  C  is  the  required  triangle. 

EXERCISES 

1.  With  sides  i   inch  and  i|  inches  long,  and  with  the 
included  angle  60°,  construct  a  triangle,  using  a  protractor  to 
lay  off  the  given  angle,  and  compasses  for  the  construction. 

2.  Taking  any  convenient  length  for  x,  as  ^  inch,  construct 
a  triangle  having  two  sides  3^  and  4X,  and  an  included  angle 
of  45°. 

3.  Construct  an  isosceles  triangle  with  one  of  the  equal 
sides  i£  inches,  and  the  vertex-angle  55°  (see  §  27). 

4.  Construct  an  isosceles  triangle  with  base  2  inches  and 
base  angles  15°. 


10 


Second-  Year  Mathematics 


5.  Construct  a  triangle  with  base  £  inch  and  base  angles 
35°  and  50°. 

17*  Geometrical  figures  that  have  the  same  shafie  are 
similar  figures. 

18.  Geometrical  figures  that  have  the  same  size,  but  not 
the  same  shape,  are  equal  or  equivalent  figures. 

19.  Geometrical  figures,  that  have  both  the  same  size  and 
the  same  shape  are  congruent  figures.     Congruent  figures  may 
be  made  to  coincide. 

20.  There  are  some  geometrical  figures,  such  as  straight 
lines,  angles,  circles,  etc.,  that  can  differ  only  in  size.     When 
such  figures  have  the  same  size  they  will  be  said  to  be  equal, 
rather  than  congruent;    since  the  question  of  shape  does  not 
arise  in  connection  with  them. 

21.  The  sign,  or  symbol,  of  similarity  is  ^.     Thus,  a  ~~  b 
means  a  is  similar  to  b. 

22.  The  symbol  of  equality,  or  equivalency,  is  =.     Thus, 
a =b  means  a  is  equal  to  b. 

23.  The   symbol   of   congruency   is  ^.     Thus,  ABC^ 
A'B'C'  means  ABC  and  A'B'C'  may  be  made  to  coincide. 

PROPOSITION  I 

24.  Theorem:    //  two  sides  and  the  included  angle  of  one  I 
triangle  are  equal,  respectively,  to  two  sides  and  the  included] 
angle  of  another  triangle,  the  triangles  are  congruent. 


e 

FIG.  14 

Hypothesis:   If  the  triangles  ABC  and  A'B'C'  (Fig.  14) 
have  A  B=A'B',  A  C=A'C',  and  angle  A  =angle  A'; 


Congruency  of  Rectilinear  Figures  and  Circles          n 


Conclusion:  Then  the  triangles  ABC  and  A'B'C'  are  con- 
gruent, i.  e.,  they  can  be  made  to  coincide. 

Proof 

Imagine  the  triangle  ABC  placed  upon  the  triangle 
A'B'C'  so  that  angle  A  shall  fit  exactly  upon  its  equal  angle, 
A',  A  B  falling  upon  A'B'  and  A  C  upon  A'C'. 

Because  A  B=A'B'  (why?),  B  will  fall  upon  B'. 
Because  A  C=A'C'  (why  ?),  C  will  fall  upon  C'. 

Then  B  C  will  fall  along  and  coincide  with  B'C',  else  there 
would  be  two  different  straight  lines  connecting  the  two  points 
B  and  C,  and  this  is  evidently  impossible. 

Hence,  the  two  triangles  are  congruent,  i.  e.,  they  coincide, 
or  fit. 

Since  ABC  and  A'B'C'  are  any  two  triangles  having  two 
sides  and  the  included  angle  of  one  equal  to  the  corresponding 
parts  of  the  other,  the  theorem  is  true. 

EXERCISES 

1.  To  find  the  distance  B  C  across  the  lake   (Fig.   15), 
C  A    was    extended    through 

A,  so  that  CA=AC'  and 
B  A  through  A  so  that 
BA=B/A.  B'C'  was  meas- 
ured and  found  to  be  £  of  a 
mile  long.  Assuming  the  two 
angles,  marked  with  arcs,  to 
be  equal,  how  long  was  B  C  ? 
Give  the  reasons  for  your  answer. 

2.  Prove:    If  AC=A'C',  B  C  =  B'C',  and    Z*  C=ZC' 
(Fig.  1  6),  the  two  triangles  are  congruent. 

3.  If  in  Fig.  16,  a=a',  b=b',   ZC=ZC',  c=8(2*-s), 


FIG.  15 


*  Z  is  the  symbol  for  angle;   Z  s,  for  angles. 


12  Second-Year  Mathematics 


and  ^Bf  =  i4s— r  +  5,  what  are  the  values  of  x,  r,  s,   Z.A, 
and  Z.B'? 


FIG.  1 6 

4.  With  sides  and  angles  designated  as  in  Fig.  1 6,  assuming 
a=a',  b=b',  and  Z.  C=Z.C',  find  x,  y,  and  z  if  .4=ioo  —  y, 

z(z— 20). 

CONSTRUCTION  VIII 

25.  Given  two  sides  and  twice  the  included  angle  of  a  tri- 
angle, to  construct  the  triangle. 

Given  the  sides  b  and  c  and  the  angle  2 A  (Fig.  17). 

b 


FIG.  17 

To  construct  the  triangle. 

The  given  angle  being  the  double  of  the  desired  angle  of 
the  triangle,  we  must  first  obtain  the  desired  angle.  This  is 
done  by  bisecting  the  given  angle,  Fig.  18  (see  FYM,  p.  147). 

Now  construct  a  triangle  as  in  Construction  VII,  with  the 
sides  b  and  c  and  the  included  angle  A  O  E. 


Congruence  of  Rectilinear  Figures  and  Circles          13 

26.  A  line  that  divides  an  angle  into  two  equal  parts  is 
called  the  bisector  of  the  angle. 

27.  The  angle  that  lies  opposite  to  the  base  of  an  isosceles 
triangle  is  the  vertex-angle  of  the  triangle. 

EXERCISES 

28.  Prove  the  following  exercises: 

1.  The  bisector  of  the  vertex- angle  of  an  isosceles  triangle 
divides  the  triangle  into  two  congruent  parts. 

2.  The  bisector  of  the  vertex-angle  of  an  isosceles  triangle 
bisects  the  base. 

3.  In  an  isosceles  triangle  the  angles  opposite  the  equal 
sides  are  equal. 

4.  All  points  of  the  perpendicular  bisector  of  a  line-segment 
are  equidistant  from  the  ends  of  the  line. 

5.  An  equilateral  triangle  is  equiangular,  i.  e.,  the  angles 
are  equal. 

29.  It  is  well  for  the  student  to  become  aware  that  we 
have  in  our  work,  thus  far,  assumed  certain  things  to  be  possible 
and  true  without  having  proved  them  to  be  so;   for  example, 
it  has  been  assumed  that  a  triangle  may  be  picked  up  and 
moved  about  in  space  without  changing  its  shape  or  size.     It 
has  also  been  assumed  that  through  any  two  points  only  one 
straight  line  can  be  drawn.     Such  fundamental  assumptions 
form  the  basis  on  which  geometry  is  built;  they  are  the  axioms 
and  postulates  of  geometry. 

30.  An  axiom  is  a  statement  the  truth  of  which  is  assumed, 
i.  e.,  accepted  without  proof. 

31.  A  postulate  is  a  statement  of  the  possibility  of  a  con- 
struction. 

An  axiom,  like  a  theorem,  has  a  hypothesis  and  a  con- 
clusion.    How  does   an   axiom  differ  from  a  theorem?    A 


14  Second- Year  Mathematics 

postulate  differ  from  a  problem?    A  theorem  differ  from  a 
problem  ? 

32.  Since  every  statement  in  a  proof  must  be  based  directly, 
or  indirectly,  upon  an  axiom,  a  postulate,  or  a  definition,  some 
of  the  most  important  axioms  and  postulates  are  now  stated. 

AXIOMS 

1.  Magnitudes  that  are  equal  to  the  same  magnitude  are 
equal  to  each  other  (comparison  axiom). 

2.  Equals  added  to  equals  give  equal  sums  (addition  axiom). 

3.  Equals  subtracted  from  equals  give  equal  differences  (sub- 
traction axiom). 

4.  Equals  multiplied  by  equal  numbers  give  equal  products 
(multiplication  axiom). 

5.  Equals  divided  by  equal  numbers  (excluding  division  by  6} 
give  equal  quotients  (division  axiom). 

6.  A  whole  is  equal  to  the  sum  of  all  its  parts. 

7.  A  whole  is  greater  than  any  of  its  parts. 

8.  The  shortest  distance  Between  two  points  is  measured  on 
the  straight  line  joining  the  points. 

9.  A  figure  may  be  moved  about,  or  turned  over  in  space, 
without  changing  its  form  or  size. 

10.  One  and  only  one  straight  line  can  join  two  points. 
Other  axioms  will  be  stated  as  they  are  needed. 

POSTULATES 

1.  A  straight  line  can  be  drawn  joining  two  points. 

2.  A  line-segment  can  be  prolonged.    * 

3.  A  circle  can  be  drawn  with  any  point  as  center  and  with 
any  radius. 

33.  The  Greek  philosopher  Plato  (429-384  B.  c.)  taught 
geometry  as  a  basis  for  the  study  of  philosophy.   He  was  accord- 


Congruency  of  Rectilinear  Figures  and  Circles          15 

ingly  much  concerned  about  its  logical  purity.  He  decided 
to  restrict  the  lines  that  could  be  used  in  plane  geometry  to 
the  straight  line  and  the  circle,  and  the  instruments  that  could 
be  used  to  'the  unmarked  straight  edge  and  the  compasses. 
Geometers  since  Plato's  time  have  conformed  to  these  restric- 
tions, as  a  convenient  means  of  denning  the  field  of  plane 
geometry. 

In  the  logic  (deductive)  of  this  book  Plato's  restrictions 
are  adhered  to.  Figures  and  construction  lines  used  in  estab- 
lishing geometrical  propositions  must  be  made  up  of  circles 
and  straight  lines  only,  i.  e.,  of  such  lines  as  can  be  drawn 
with  the  unmarked  ruler  and  the  ungraduated  circle.  For 
the  deductive  proofs  neither  the  markings  on  a  foot-rule  or  a 
yard-stick,  nor  on  a  protractor  can  be  used. 

But  another  important  part  of  geometrical  study  is  to  learn 
how  to  find  out  things,  how  to  discover  new  truths,  i.  e.,  how 
to  find  out  what  is  likely  to  be  true  and  to  state  it  in  form 
for  logical  proof  or  disproof.  In  this  work  of  geometrical 
exploration  and  discovery  any  sort  of  line  or  instrument  that 
furnishes  a  clue,  affords  a  hint,  or  gives  any  sort  of  real  help, 
rational  or  intuitional,  will  be  freely  admitted.  The  graduated 
measuring-stick,  the  protractor,  cross-lined  paper,  graph, 
logarithmic  tables,  algebraic  equations,  etc.,  are  all  welcomed 
here  as  helps  to  inductive  study  and  research. 

CONSTRUCTION  IX 

34.  To  draw  a  triangle  having  given  two  angles  and  the 
side  included  by  the  vertices  of  these  angles. 

Given  the  angles  x  and  y  (Fig.  1 9)  and  the  side  a. 

To  construct  a  triangle  having  the  angles  x  and  y  and  the 
side  a,  included  between  their  vertices. 

On  the  indefinite  straight  line,  A  X,  lay  off  A  B=a. 


1 6  Second-Year  Mathematics 

At  A  make  an  angle  equal  to  x,  and  at  B  an  angle  equal  to  y 
(see  Construction  VI). 

Prolong  the  sides  of  the  constructed  angles  till  they  meet, 
as  at  C. 

a 


A  B  C  is  the  required  triangle. 

Does  it  seem  possible  to  construct  a  triangle  of  any  other 
shape  or  size,  using  the  parts  a,  x,  and  y  of  Fig.  19? 


PROPOSITION  II 


35.  Theorem:  //  two  triangles  have  two  angles  and  the 
side  included  by  their  vertices  in  the  one,  equal,  respectively,  to 
the  corresponding  parts  in  the  other,  the  triangles  are  congruent. 

Hypothesis:      If    the    two    triangles    ABC   and  A'B'C' 
(Fig.  20)  have  Z£=Z£',  ZC=ZC',  and  a = a'; 
Conclusion:  The  two  triangles  are  congruent. 

Imagine  the  triangle  A  B  C  so  placed  upon  the  triangle 
A'B'C'  (turned  over  if  necessary)  that  side  a  coincides  with 
side  a',  the  vertices  B  and  C  falling  on  the  vertices  B'  and  C', 
respectively. 


Congruency  of  Rectilinear  Figures  and  Circles  17 

Let  A  fall  on  the  same  side  of  a'  as  A'. 

Because   Z-B=Z#'  (why?),  c  will  fall  along  c',  and  the 
vertex  A  will  fall  somewhere  on  c'  or  its  extension. 

Because  ZC=ZC'  (why?),  b  will  fall  along  b',  and  the 

vertex  A  will  fall  somewhere  on  b'  or  its  extension.  • 


Because  A  falls  at  the  same  time  upon  c'  and  upon  b',  it 
must  fall  at  the  only  point  c'  and  b'  have  in  common,  viz.,  at 
their  point  of  intersection,  A'. 

Hence,  the  triangles  ABC  and  A'B'C'  coincide,  and  are 
congruent.  But  ABC  and  A'B'C'  are  any  two  triangles 
having  two  angles  and  the  intervening  side  of  the  one  equal  to 
the  corresponding  parts  of  the  other. 

The  theorem  is  therefore  true. 

EXERCISES 

1.  The  perpendicular  bisector  of  the  base  of  an  isosceles 
triangle  passes  through  the  opposite  vertex  (use  Proposition  II, 
p.  16). 

Assume  for  the  present  that  all  right  angles  are  equal. 

2.  In  Fig.  21  the  distances  S  A  and  S  B  of  a  lighthouse,  S, 
from  the  points  A  and  B  are  desired. 

The  angle  A  B  S'  is  made  equal  to  the  angle  A  B  S,  and 
B  A  S'  is  made  equal  to  B  A  S.  A  S'  was  measured  and  found 


i8 


Second-Year  Mathematics 


to  be  2,680  ft.  and  B  S'  was  found  to  be  3,420  ft.     How  far 
is  S  from  A  and  B  ? 

3.  The  bisector 
of  the  vertex-angle 
of  -     an      isosceles 
triangle  is  perpen- 
dicular to  the  base 
(see     Exercise      2, 
P-  13)- 

4.  Oblique  lines 
drawn  to  a  line  from 
a  point   on  a  per- 
pendicular    to    the 
line     and     makin 


FlG-  2I 


equal  angles  with  the  perpendicular  are  equal. 

5.  Oblique  lines,  drawn  from  a  point  on  a  perpendicular  to 
a  line  and  making  equal  angles  with  the  line,  are  equal. 

Assume  that  the  sum  of  the  angles  of  any  triangle  is  180°  (see  p.  31,  3). 

6.  An  equiangular  triangle  is  equilateral. 

An  equiangular  triangle  is  a  triangle  all  of  whose  angles  are  equal 
(see  Exercise  5,  p.  13). 

7.  If  two  angles  of  a  triangle  are  equal,  the  sides  opposite 
the  equal  angles  are  equal,  and  the  triangle  is  isosceles  (see 
FYM,  p.  344,  9)- 

c  c' 


A         c  BA       c/ 

FIG.  22 

8.  In  triangles   marked  as  in   Fig.    22,   supposing  b  =  b', 
—  A'  r" r'  n  — 

—  JT!    ,    \^  — \^    ,    (I — i 


J    T   M    2 

and  B'  =  —      —  .     Find  x,  y,  n,  a,  c,  and  B'. 


Congruency  of  Rectilinear  Figures  and  Circles          ig 

PROPOSITION  III 

36.  Theorem:  If  the  three  sides  of  one  triangle  are  equal, 
respectively,  to  the  three  sides  of  another  triangle,  the  triangles 
are  congruent. 

Given,  in  Fig.  23,  A  B=A'B',  A  C=A'C',  and  B  C=B'C'. 


A  B  A  B' 

FIG.  23 

To  prove:    A  B  C  *=  A'B'C'. 

Place  A  B  on  A'B'  so  that  A  falls  upon  A'  and  B  upon  B'. 

Let  C  fall  on  the  same  side  of  A'B'  as  C'. 

Then  if  C  does  not  fall  on  C'  suppose  it  to  fall  at  C",  some 
other  point. 

But  A'C'=A'C"  (why?)  and  B'C'=B'C"  (why?).  Hence 
both  A'C'C"  and  B'C'C"  are  isosceles  triangles. 

A  perpendicular  to  C'C"  at  its  middle  point  passes  through 
both  A'  and  B'  (see  Exercise  i,  p.  17). 

This  is  impossible  unless  the  middle  point  of  C'C"  lies 
on  A'B'  (see  Axiom  10).  The  middle  point  of  C'C"  cannot 
lie  on  A'B'  without  destroying  the  triangles,  since  C  and  C' 
were  made  to  fall  on  the  same  side  of  A'B'. 

Therefore  the  supposition  that  C  does  not  fall  on  C'  is 
false,  and  the  triangles  ABC  and  A'B'C'  are  congruent. 

But  ABC  and  A'B'C'  are  any  two  triangles  having  the 
three  sides  of  the  one  equal,  each  to  each,  to  the  three  sides  of 
the  other,  and  the  theorem  is  then  true. 


2o  Second-Year  Mathematics 

This  is  an  example  of  indirect  proof.  It  consists  in  assum- 
ing the  opposite  or  the  negative  of  what  is  to  be  proved,  and 
showing  the  assumption  to  lead  to  an  absurdity.  Hence 
it  is  called  a  "reductio  ad  absurdum." 

EXERCISES 

1.  A  line  drawn  from  the  middle  point  of  the  base  to  the 
opposite  vertex  of  an  isosceles  triangle  is  perpendicular  to  the 
base. 

2.  Oblique  line-segments  drawn  from  a  point  in  a  perpen- 
dicular to  a  line  and  cutting  off  equal  distances  from  the  foot 
of  the  perpendicular  are  equal. 

3.  Equal  oblique  line-segments  drawn  from  a  point  in  a 
perpendicular  to  a  line,  meet  the  line  at  points  equally  distant 
from  the  foot  of  the  perpendicular  (see  §  28,  3) . 

37.  The  point  of  intersection  of  a  line  with  a  perpendicular 
is  called  the  foot  of  the  perpendicular. 

38.  In  congruent  triangles,   corresponding  sides  are  sides 
that  lie  opposite  equal  angles. 

Corresponding  angles  are  angles  that  lie  opposite  equal 
sides. 

39.  The  sides  and  angles  of  a  triangle,  or  other  figure,  are 
called  the  parts  of  the  triangle,  or  figure. 

0' 


i .  With  the  parts  of  two  triangles  designated  as  in  Fig.  24, 
and  with  a=ar,  b=b',  c=c',  find  r,  s,  t,  and  all  the  other  angles 
under  the  following  conditions: 


Congruency  of  Rectilinear  Figures  and  Circles          21 

Angle  A  Angle  A'         Angle  B  Angle  B'      Angle  C       Angle  C' 

(1)  5(r+s)        2(29-5)     2(5+0         86-/  4^  +  2/        2(9+*) 

(2)  lr+$s  +  ±t      75-5    ^+£5+43  90-5  ^+£5+64  195-* 

2.  Supposing  the  sizes  of  the  three  angles  of  an  equilateral 
triangle  may  be  represented  by  the  numbers,  r  +  2s,  $t  —  ior, 
and  s+t,  respectively,  and  that  the  sum  of  all  three  is  180°, 
find  r,  s,  and  J,  and  each  of  the  angles. 

40.  Propositions  I,  II,  and  III  are  the  most  fundamental, 
if  not  the  most  important,  theorems  of  plane  geometry.  The 
proofs  of  all  three  as  given  here  furnish  examples  of  the  method 
of  establishing  congruency  by  so  superposing  one  figure  upon 
another  as  to  make  the  figures  coincide. 

The  proof  of  Proposition  III  is  of  this  superposition  type 
mixed  with  the  "reductio  ad  absurdum"  type.  These  three 
proofs  should  be  thoroughly  mastered. 


AXIOM  ii 
All  straight  angles  are  equal. 

PROPOSITION  IV 

41.  Theorem:     All  right  angles  are  equal. 

To  prove  this  use  Axiom  u  just  given  and  Axiom  5,  p.  14. 

42.  The  supplement  of  an  angle  is  the  difference  obtained 
by  subtracting  it  from  a  straight  angle  (180°). 

43.  The  complement  of  an  angle  is  the  difference  obtained 
by  subtracting  it  from  a  right  angle  (90°). 

PROPOSITION  V 

44.  Theorem :     The  supplements  of  equal  angles  are  equal. 
Use  Axioms  n  and  3. 

PROPOSITION  VI 

45.  Theorem:     The  complements  of  equal  angles  are  equal. 
The  proof  is  left  to  the  student. 


22 


Second-  Year  Mathematics 


46.  Two  angles  having  a  common  vertex  and  a  common  side 
between  them  are  adjacent  angles. 


EXERCISES 

i.  The  sum  of  the  adjacent  angles  formed  by  one  line 
meeting  another  is  a  straight  angle. 

This  is  evident  from  Axiom  6,  p.  14. 

^/2?)  One  of  two  complementary  angles  is  x  and  the  other 
is  x2.     Find  x  and  the  two  angles. 

3.  One  of  two  complementary  angles  is  6y  and  the  other 
is  4(^  +  5)-     Find  y  and  the  angles. 

4.  One  of  two  supplementary  angles  may  be  designated  y2 
and  the  other  8y.     Find  y  and  the  angles. 

(Q  The  angles  x  and  y  are  complementary  and  their  differ- 
ence is  10°.     Find  x  and  y. 

6.  The  angles  r  and  35  are  supplementary  and  r— 5  =  20°. 
Find  r,  s,  and  35-. 

7.  The  angles  that  lie  opposite  the  equal  sides  in  an  isos- 
celes triangle  are  represented  in  magnitude  by  7^+8  and 
80  —  $x,  respectively.     Find  x  and  the  two  angles. 

8.  If  one  of  two  adjacent  angles  formed  by  one  line  meeting 
another  is  a  right  angle,  the  other  is  also  a  right  angle.     Prove. 

9.  If  two  straight  lines  intersect  (cross)  one  another,  and 
one  of  the  angles  formed  is  a  right  angle,  the  other  three  are 
also  right  angles.     Prove. 

10.  The  sum  of  all  the  angles  formed  about  a  point  on  one 
side  of  a  line  is  a  straight  angle.     Prove. 

11.  The  sum  of  all  the  angles  formed  about  a  point  is  two 
straight  angles,  or  four  right  angles.     Prove. 

PROPOSITION  VII 

47.  Theorem:     If  the  sum  of  two  adjacent  angles  is  a 
straight  angle,  their  exterior  sides  form  a  straight  line. 

This  follows  directly  from  the  definition  of  straight  angle,  §  15. 


Congruence  of  Rectilinear  Figures  and  Circles  23 

EXERCISES 

1.  The  bisectors  of  two  supplementary  adjacent  angles  are 
perpendicular  to  each  other.     Prove. 

When  two  lines  intersect  each 
other  at  right  angles,  either  line  is 
said  to  be  perpendicular  to  the  other. 

2.  The  angle  ABC  (Fig.  25) 
is  a  right  angle.     D  B  E  is  a 
straight  line.     To  what  is  the 
sum,  x+y,  equal? 

The  opposite  angles,  as  a  and  c, 
or  b  and  d  (Fig.  26)  formed  by  two 
intersecting  lines  are  called  vertical 
angles. 

3.  Prove  that  if  two  lines 
intersect,  the  vertical  angles  are 
equal. 

4.  The  bisectors  of  two  op- 
posite, or  vertical,  angles  form  a  straight  line. 

5.  The  bisector  of  an  angle  bisects  also  its  vertical  angle. 

PROPOSITION  VIII 

48.  Problem:    From  a  point  without  a  line  to  draw  a  per- 
pendicular to  the  line. 

See  Fig.  27,  and  FYM,  p.  146. 


FIG.  26 


FIG.  27 


Second-Year  Mathematics 


PROPOSITION  IX 

49.  Theorem:    Only  one  perpendicular  can  be  drawn  from 
a  point  without  a  line  to  the  line. 

Let  P  O  be  a  perpendicular  from  P  to  A  B  (Fig.  28). 
To  prove  P  O  the  only  perpendicular  from  P  to  A  B. 

Proof:  If  there  can  be  another 

T-\ 

perpendicular,  let  it  be  P  C.  Pro- 
long P  O  to  P7,  making  O  P7=P  O. 

Join  C  and  P7  and  prove  P  C  O 
and  P7C  O  congruent  triangles. 
~£  P  C,  being  a  perpendicular  to  A  B 
(by  supposition),  P  C  P7  is  a  straight 
line  (Proposition  VII). 

If  then  P  C  could  be  any  other 
perpendicular  than  P  O,  there  would 
be  two  straight  lines  connecting  P 
and  P'.  This  is  impossible.  Why?  (See  Axiom  10,  p.  14.) 

Hence,  there  can  be  no  other  perpendicular  from  P  to  A  B 
than  P  O. 

50.  The  symbol  for  perpendicularity  is  1. 

PROPOSITION  X 

51.  Problem:   From  a  point  on  a  line,  to  draw  a  perpendicu- 
lar to  the  line. 

See  Fig.  29,  and  FYM,  p.  145. 


o 


FIG.  28 


P 

FIG.  29 


1T23 


Congruency  of  Rectilinear  Figures  and  Circles          25 


PROPOSITION  XI 

52.  Theorem:  If  each  of  two  points  of  one  line  is  equally 
distant  from  two  points  of  another  line,  the  lines  are  per- 
pendicular. 

Let  P  and  Q  of  either  Fig.  30  or  Fig.  31  be  each  equally 
distant  from  the  ends  A  and  B  of  the  line  A  B. 


V 


A" 


" 


-v   A 

FIG.  30 


\   \ 


0 


A' 


^ 


N 


K? 

\ 


FIG.  31 

To  prove  the  lines  P  Q  and  A  B  perpendicular. 
Proof:    Connect  P  with  A  and  B,  and  Q  with  A  and  B. 
Prove  triangles  P  A  Q  and  P  B  Q  congruent. 
Prove  triangle  A  O  Q  congruent  to  triangle  B  O  Q. 
Then  ZQ  O  A=  ZQ  O  B,  and  the  lines  P  Q  and  A  B  are 
perpendicular.     Why  ? 

PARALLELS  AND  PARALLELOGRAMS 

53.  Parallel  lines  are  lines  which  lie  in  the  same  plane  and 
do  not  intersect  however  far  prolonged  in  both  directions. 
The  symbol  for  parallelism  is  || . 


26 


Second-Year  Mathematics 


AXIOM  12 

One  and  only  one  parallel  to  a  line  can  be  drawn  through 
a  point  outside  of  the  line. 

PROPOSITION  XII 

/      54.  Theorem:    //  each  of  two  lines  is  parallel  to  a  third 
\line,Jhcy  are  parallel  to  each  other. 

Given  A  B  and  C  D  each  parallel  to  E  F. 


A- 
C- 


-F 


FIG.  32 


To  prove  A  B  |   CD. 
The   lines   AB   and  CD 
(Fig.  32)  are  supposed  to  lie 
in  the  plane  of  the  page. 

Suppose  A  B  could  intersec  t 
C  D,  through  some  point  as  P. 
Then,  through  P  there  would  be   two  parallels  to  E  F. 
Why?    This  is  impossible.     Why?     (See  Axiom  12.) 

A  B  and  C  D  are  then  lines  which  lie  in  the  same  plane 
and  do  not  intersect.     Hence,  A  B  and  C  D  are  parallel. 

This  furnishes   another   example  of  the   "reductio   ad  absurdum" 
type  of  proof. 

PROPOSITION  XIII 

55.  Theorem:     Two  lines  that  are  perpendicular  to  the 
.same  line  are  parallel  to  each  other. 

Let  AB  and  CD  both 
be  J_  E  F  (Fig.  33). 

To  prove  A  B   and    A. 
C  D  parallel. 

A  B    and    C  D    are 
supposed  to  lie  in  the    C 
plane  of  the  page. 

Suppose  A  B  and 
C  D  could  intersect  at 
some  point  as  P.  FIG.  33 


Congruency  of  Rectilinear  Figures  and  Circles          27 


Then  through  P  there  would  be  two  perpendiculars  to 
E  F,  which  is  impossible.  Why  ? 

/.  A  B  and  C  D  do  not  intersect. 

Since  A  B  and  C  D  lie  in  the  same  plane  and  do  not  inter- 
sect, they  are  parallel.  Why? 


56.  A  line  cutting  two  or  more  lines  is  called  a  transversal. 


Thus  E  F  (Fig.  34),  is  a  transversal. 

The  angles  a,  b,  c',  and  d'  are  ex- 
terior angles. 

The  angles  c,  d,  a',  and  b'  are  in- 
terior angles. 

The  angles  a  and  c',  b  and  d',  a' 
and  c,  d  and  b'  are  alternate  angles. 

The  angles  a  and  a',  b  and  b',  c 
and  c',  d  and  d'  are  corresponding 
angles. 


r 


FIG.  34 


The  angles  c  and  a',  b'  and  d  are  alternate-interior  angles. 
The  angles  b  and  d',  a  and  c'  are  alternate -exterior  angles. 


PROPOSITION  XI1 


/      57.  Theorem:   If  two  alternate-interior  angles,  formed  by  two 
I  Urns  and  a  transversal,  are  equal,  the  two  lines  are  parallel. 

Let  the  angles  a 
and  a'  (Fig.  35) ,  made 
by  A  B  and  C  D  with 
the  transversal  E  F, 
be  equal. 

To  prove  A  B  and 
C  D  parallel. 

The  lines  are  all 
supposed  to  lie  in  the 
plane  of  the  page. 


A                          K       HX            r> 

f\ 

£ 

L 


FIG.  35 


28 


Second-  Year  Mathematics 


Through  the  middle  point,  M,  of  G  H  draw  a  perpendicular 
to  A  B,  and  prolong  it  to  meet  C  D  at  L. 

What  parts  of  triangles  M  G  L  and  M  H  K  are  equal  ? 

Prove  M  G  L  and  M  H  K  congruent. 

M  K  H  is  a  right  angle.     Why  ? 

Therefore  M  L  G  is  a  right  angle.     Why  ? 

A  B  and  C  D  are  both  perpendicular  to  K  L.     Why  ? 

Therefore  A  B  ||  C  D.     Why? 

EXERCISE 

i.  If  one  of  two  parallel  lines  is  perpendicular  to  a  third 
line,  the  other  is  also.  Prove. 

A  corollary  is  a  theorem,  the  proof  of  which  is  contained 
in  the  proof  of  other  propositions. 

58.  Corollary:  If  two  lines  are  cut  by  a  transversal  in  such 
a  way  that  the  sum  of  two  interior  angles  on  the  same  side  of  the 
transversal  is  equal  to  a  straight  angle,  the  two  lines  are  parallel. 

The  arrangement  of  the  proof  is  left  to  the  student. 

PROPOSITION  XV 

Theorem :  If  two  lines  cut  by  a  transversal  are  parallel, 
alternate-interior  angles  are  equal. 

Let  A  B  and 
C  D  be  two  paral- 
lel lines,  cut  by 
the  transversal 
E  F  (Fig.  36). 

To  prove  angle 
A  H  G  equal  to 
angle  HGD. 

Proof:  At  H 
on  the  line  E  F, 
ZHGD. 


_----N 


NT 
r       / 

~7H 

& 

T) 

Y 

an   angle  can 

FIG.  36 
be  drawn 

'  .£/ 

equal   to 

Congruency  of  Rectilinear  Figures  and  Circles         29 


If  A  B  does  not  make  angle  A  H  G  equal  to  H  G  D,  suppose 
some  other  line,  as  M  H  N  does.  Then  M  H  G  =H  G  D. 

Then  M  N  and  C  D  are  parallel.    Why  ? 

But  A  B  is  assumed  parallel  to  C  D. 

Hence,  through  H  there  would  be  two  parallels  to  C  D, 
which  is  impossible.  Why?  (Axiom  12,  p.  26.) 

Therefore  no  other  line  than  A  B  through  H  can  make 
with  E  F  and  C  D  the  alternate-interior  angles  equal. 

Hence,  AH  G=HGD. 

This  is  another  example  of  the  indirect  form  of  proof,  known  as 
" reductio  ad  absurdum." 

60.  One  theorem  is  said  to  be  the  converse  of  another, 
when  the  hypothesis  and  the  conclusion  of  the  one  are,  respec- 
tively, the  conclusion  and  the  hypothesis  of  the  other. 

Give  examples  of  theorems  and  their  converse;  of  axioms 
and  their  converse. 

The  converse  of  a  theorem  is  not  true,  simply  because  the 
theorem  is  true.     Some  converses  are  true  and  some  are  not. 
A  proof  is  necessary  before  they  can  be  accepted  as  true. 
.  Give  cases  in  which  the  converse  of  a  theorem  is  not  true. 

('EXERCISES 
.  Supposing 'A  B  ||  ClT(Fig.  37)  prove  the  following: 

(2)  b  =  V          (4)  d=d'         (6)  b=d'         (8)  b'=d. 


FIG.  37 


30 


Second-Year  Mathematics 


2.  Prove  also  the  following: 

(1)  a  and  b'  are  supplementary 

(2)  a'  and  b  are  supplementary 

(3)  c'  and  d  are  supplementary 

(4)  c  and  d'  are  supplementary. 

3.  A  perpendicular  to  one  of  two  parallels  is  perpendicular 
to  the  other  also.     Prove. 

4.  The  rails  of  two  railroad  tracks  cross  so  that  one  of  the 
1  6  angles  formed  is  36°.     Make  a  sketch  and  state  the  value 
of  each  of  the  other  1  5  angles. 

5.  If  the  sum  of  the  interior  angles  on  one  side  of  a  trans- 
versal to  two  lines  is  less  than  a  straight  angle  the  lines  meet. 

61.  Suppose  an  observer  standing  at  the  vertex  of  an  angle 
as  at  O  (Fig.  38),  and  looking  off  over  the  angular  space  in 
the  direction  O  D,  one  side  of  the  angle  lies  on  his  right  and 
the  other  on  his  left. 


J> 


0 


FIG.  38 


FIG.  39 


An  angle  may  thus  be  regarded  as  having  a  right  side  and 
a  left  side  (see  Figs.  38  and  39). 

EXERCISES 

i  .  If  the  sides  of  two  angles  are  parallel,  right  to  right  and 
left  to  left  (Fig.  40),  the  angles  are  equal.  Prove. 

2.  If  the  sides  of  two  angles  are  parallel,  right  to  left  and 
left  to  right  (Fig.  41),  the  angles  are  supplementary.  Prove. 


Congruency  of  Rectilinear  Figures  and  Circles          31 

3.  The  sum  of  the  three  angles  of  any  triangle  is  equal  to 
a  straight  angle,  or  to  two  right  angles.  Prove.  (See  Fig.  42. 
See  also  FYM,  pp.  54,  55.) 


62.  A  triangle  that  contains  a  right  angle  is  a  right  tri- 
angle. 


FIG.  41 

63.  A  triangle  that  contains  an  obtuse  angle  is  an  obtuse 
triangle. 


FIG.  42 

64.  A  triangle,  all  of  whose  angles  are  acute,  is  an  acute 
triangle. 


32 


Second-Year  Mathematics 


EXERCISES 

1.  How  many  obtuse  angles,  at  most,  may  a  triangle  have  ? 
How  many  right  angles,  at  most?    How  many  acute  angles; 
at  most?    How  many  acute  angles,  at  least?     Give  reasons 
for  your  answers. 

2.  The  sum  of  the  acute  angles  of  a  right  triangle  is  a  right 

angle,  i.  e.,  the  acute  angles  of  a  right 
triangle  are  complementary. 

3.  If    the  sides   of  two   angles 


I 


I 


FIG.  43 


FIG.  44 


are  perpendicular,  right  to  right  and   left  to  left  (Fig.  43), 
the  angles  are  equal. 

4.  If  the  sides  of  two  angles  are  perpendicular,  right  to 
left  and  left  to  right  (Fig.  44),  the  angles  are  supplementary. 

CONSTRUCTION  X 

65.  Problem :    To  draw  a  line  parallel  to  a  given  line  through 
a  given  point. 

pi  ^  -p  I«et  P  be  the  given 

point  and  A  B   be  the 
given  line  (Fig.  45). 

i \t To  draw  a  parallel  to 

X  J3     A  B  through  P. 

With  P  as  a  center 


\ 


\ 
\ 


FIG.  45 


and  with  a  radius  longer  than  the  distance  from  P  to  A  B,  draw 
an  arc  as  C  F. 


Congruency  of  Rectilinear  Figures  and  Circles         33 


With  C  as  a  center  and  with  the  same  radius,  draw  the 
arcPE. 

With  a  radius  equal  to  the  distance  from  E  to  P  and  with 
C  as  a  center,  draw  an  arc  as  at  F  across  arc  C  F. 

Join  F  and  P.     The  line  F  P  is  the  required  parallel. 

F  P  will  later  be  proved  parallel  to  A  B. 

66.  A  plane  figure  bounded  by  four  straight  lines  is  a 
quadrilateral. 

67.  A  quadrilateral  having  two  pairs  of  parallel  sides  is  a 
parallelogram. 

EXERCISES 

1.  The  sum  of  the  four  angles  of  a  quadrilateral  is  two 
straight  angles  (see  Exercise  3,  §  61, 

and  Fig.  46). 

The  angles  marked  with  arcs  (Fig.  47) 
are  exterior  angles  (see  FYM,  p.  55, 
§39). 

2.  The  sum  of  the  exterior  angles 
of  a  triangle   is  four  right   angles 
(see  FYM,  §  40,  Exercise  4). 


FIG.  46 


FIG.  47 

3.  The  sum  of  the  exterior  angles  of  a  quadrilateral  is  four 
right  angles  (see  Fig.  47). 


34 


Second-Year  Mathematics 


PROPOSITION  XVI 

68.  Problem:    To  construct  a  parallelogram,  having  given 
two  adjacent  sides  and  the  included  angle. 

Let  a  and  b  be  the  two  adjacent  sides,  and  let  x  be  the 
included  angle  (Fig.  48). 

a  To  construct  the  parallelogram. 

I  Lay  off  from  the  vertex  of  the  angle 

upon  the  sides  (prolonged  if 
/~       necessary)  the  sides  a  and  b, 
~~f  as  A  Band  A  B. 

/  Through  D  draw  a  parallel 

to  A  B  and  through  B  draw 
a  parallel  to  A  D. 

Let   C   denote    the   inter- 
secting point  of  the  parallels. 
Then  A  B  C  D  is  the  required  parallelogram.     Why  ? 

EXERCISES 

1.  Prove,  that  if  two  adjacent  sides  and  the  included  angle 
of  one  parallelogram  are  equal,  respectively,  to  the  correspond- 
ing parts  of  another,  the  parallelograms  are  congruent. 

2.  Prove,  the  adjacent  angles  of  a  parallelogram  are  supple- 
mentary. 

3.  Prove,  the  opposite  angles  of  a  parallelogram  are  equal. 


PROPOSITION  XVII 


b 

FIG.  48 


69.  Theorem:    The  op- 

posite  sides  of  a  parallelo- 
gram are  equal. 


p.  334,  Problem  7. 


FYM, 


FIG.  49 


Congruency  of  Rectilinear  Figures  and  Circles          35 


PROPOSITION  XVIII 


^^*"~  .  ~^  • '    ^^ 

•    70.  Theorem:    If  the  opposite  sides  of  a  quadrilateral  are 
equal,  the  quadrilateral  is  a  parallelogram.  , 

Converse  of  Proposition  XVII.  See  p.  19,  §  36  and  Proposition 
XIV. 

Let  ABCD  be  a  quadrilateral  having  AB=DC,  and 
AD=BC(Fig.  50). 

To   prove  A  B  C  D    a  7) /*r 

~"  |/>       «•  7  O 

parallelogram. 

Draw  a  diagonal,  as  A  C. 

Prove      the      triangles         A 
ABC    and    ADC    con- 

tlG.  50 

gruent. 

Then  since  y=y'  (why?)  and  since  A  C  is  a  transversal  to 
A  Band  DC,  AB||DC.  Why? 

Also,  since  x=x',  and  A  C  is  also  a  transversal  to  A  D 
andBC,  A  D  ||  B  C.  Why? 

The  quadrilateral  is  then  a  parallelogram.     Why  ? 

PROPOSITION  XIX 

71.  Theorem:  If  the  opposite  angles  of  a  quadrilateral  are 
equal,  the  quadrilateral  is  a  parallelogram. 


B 

FIG.  51 

Let  ABCD  be  a  quadrilateral    having  a=c,  and  b—d 
(Fig-  50- 

To  prove  the  quadrilateral  a  parallelogram. 


36  Second-Year  Mathematics 

Proof 

a+b+c+d=4  right  angles.     Why? 

But  c=a  and  b=d.     Why? 

Therefore  a+d  =  a  straight  angle,  or  two  right  angles. 
Why? 

Regarding  A  D  as  a  transversal  to  A  B  and  DC,  A  B  1 1 
D  C  (see  §  58). 

Also  d+c=a.  straight  angle  or  two  right  angles. 

Therefore  AD  ||  BC.     Why? 

And  A  B  C  D  is_a  parallelogram. 

PROPOSITION  XX 

72 .  Theorem :     The  diagonals   of  a   parallelogram  bisect  I 
each  other: 

Given  the  parallelogram  with  the  diagonals  A  C  and  B  D 
(Fig-  52). 


& 

FIG.  52 

To  prove  A  C  and  B  D  bisect  each  other. 

Proof 

Call  the  point  of  intersection  of  the  diagonals  E. 
What  parts  of  the  triangle  A  B  E  are  equal  to  corresponding 
parts  of  D  C  E  ? 

Prove  the  triangles  ABE  and  D  C  E  congruent. 
D  E  then  equals  E  B,  and  A  E  equals  E  C.     Why  ? 
See  FYM,  p.  334,  Problem  8. 


Congruency  of  Rectilinear  Figures  and  Circles         37 

EXERCISES 

1.  Prove,  conversely,  if  the  diagonals  of  a  quadrilateral 
bisect  each  other,  the  quadrilateral  is  a  parallelogram  (use 

Fig-  52)- 

2.  Prove,  a  diagonal  divides  a  parallelogram  into  two  con- 
gruent triangles  (see  FYM,  p.  345). 

PROPOSITION  XXI 

73.  Theorem:    If  two  sides  of  a  quadrilateral  are  equal  and 
parallel,  the  quadrilateral  is  a  parallelogram. 

Given     the     quadrilateral  J) p 

A  B  C  D     with    A  D     equal 
and  parallel  to  B  C  (Fig.  53). 

To  prove  A  B  C  D  a  paral- 
lelogram. 

Draw   the   diagonals  A  C  FIG.  53 

and  B  D ;  let  E  be  their  intersection. 

Regarding  D  B  as  a  transversal  to  A  D  and  C  B,  x=xf. 

Why? 

Regarding  A  C  as  a  transversal  to  A  D  and  C  B,  y=y'. 
Why? 

Prove  A  E=E  C  and  D  E=E  B. 

A  B  C  D  is  then  a  parallelogram  (see  Exercise  i  above), 
i  . 1 ... 

74.  A  rectangle  is  a  parallelogram  one  of 
•whose  angles  is  a  right  angle. 

EXERCISES 

1.  Prove,  that  if  one  of  the  angles  of  a  parallelogram  is  a 
right  angle,  all  the  angles  are  right  angles. 

2.  Prove,  the  diagonals  of  a  rectangle  are  equal. 

75.  A   square   is   a   rectangle  two  of.  whose 
adjacent  sides  are  equal. 


38  Second-Year  Mathematics 

EXERCISES 

1.  Prove,  that  if  two  adjacent  sides  of  a  rectangle  are  equal, 
all  the  sides  are  equal. 

2.  Prove,  the  diagonals  of  a  square 
bisect  each  other  perpendicularly. 

76.  A  rhombus  is  a  parallelogram  two 
of  whose  adjacent  sides  are  equal. 

EXERCISES 

1.  Prove,  that  if  two  adjacent  sides  of  a  parallelogram  are 
equal,  all  the  sides  are  equal. 

2.  Prove,  the  diagonals  of  a  rhombus  bisect  each  other  per- 
pendicularly (see  FYM,  p.  346,  Exercise  6). 

77.  A  trapezoid  is  a  quadrilateral  having      /                \ 
two  parallel  sides.  L A 

The  parallel  sides  of  a  trapezoid  are  the  bases. 
A   trapezoid  whose    non-parallel   sides   are    equal    is    an 
isosceles  trapezoid. 

A          ,  .,  ,      ,  ,      .      (  two  pairs  of  parallel  sides  is  a  parallelogram 
A  quadrilateral  having  j  Qne  ££  of  J*~^|  s;des  fa  a  £apezoid5 

.  r.     •      (  one  right  angle  is  a  rectangle 

A  parallelogram  having  J  ^  Jjacenf  sides  equal  |  a  rhombus 

A  parallelogram  that  is  both  a  rectangle  and  a  rhombus  is  a  square. 

,  (  parallelogram   \  r£ctangle  {  square 
Quadrilateral 


1 

EXERCISES 

i.  Prove,  that  if  the  two  angles  at  the  ends  of  one  base  of 
a  trapezoid  are  equal,  the  trape- 
zoid is  isosceles. 

Draw     C  E  <\  D  A.      Prove     C  B 
=  D  A  (Fig.  54). 

2.  Prove,   conversely,  if    the 
FIG.  54  non-parallel  sides  of  a  trapezoid, 


Congmency  of  Rectilinear  Figures  and  Circles         39 

that  is  not  a  parallelogram,  are  equal,  the  angles  at  the  ends 
of  one  base  are  equal. 

PROBLEMS 

1.  Given  the  base  and  one  base  angle  of  an  isosceles  tri- 
angle, to  construct  the  triangle. 

2.  Given  one  angle,  one-half  of  another  angle,  and  the  side 
included  by  the  vertices  of  these  two  angles,  to  construct  the 
triangle. 

To  solve  this  it  is  necessary  to   construct  an  angle    equal  to  the 
double  of  a  given  angle. 

Let  a  be  the  given  half-angle  (Fig.  55). 

To  construct  an  angle  that  is  the  double 
of  a. 

With  a  convenient  radius  draw  the  arcs 
M  N  and  B  D,  the  former  with  O  as  center 
and  the  latter  with  A  as  center. 

With  the  distance  M  N  as  radius  and 
with  B  as  center  draw  an  arc  at  C.  _^____ 

Then  with  C  as  center  and  with  the  same        *t  •&  * 

radius,'  draw  the  arc  at  D.     Join  A  D.  FIG.  55 

B  A  D  is  then  the  double  of  a.     Why  ? 

Now  solve  Problem  2. 

3.  Construct  an  angle  equal  to  three  times  a  given  angle. 

4.  Construct  an  angle  equal  to  the  sum  of  two  given  angles. 

5.  Construct  an  angle  equal  to  the  difference  of  two  given 
angles. 

6.  Given  two  adjacent  sides  of  a  rectangle,  to  construct 
the  rectangle. 

7.  Given  a  side  of  a  square,  to  construct  the  square. 

8.  Given  three  consecutive  sides   and  the  two  included 
angles  of  a  quadrilateral,  to  construct  the  quadrilateral. 

9.  Given  a  diagonal  of  a  square  to  construct  the  square. 


Second-Year  Mathematics 


10.  Given  the  diagonals  of  a  rhombus,  to  construct  the 
rhombus. 

11.  To  construct  the  complement  of  a  given  angle. 

12.  To  construct  the  supplement  of  a  given  angle. 

13.  Given  the  two  sides  including  the  right  angle  of  a  right 
triangle,  to  construct  the  triangle. 

EXERCISES 
78.  Prove  the  following  theorems: 

1.  If  the  bisector  of  any  angle  of  a  triangle  is  perpendicular 
to  the  opposite  side,  the  triangle  is  isosceles. 

2.  If  the  middle  points  of  the  equal  sides  of  an  isosceles 
triangle  are  joined  to  the  opposite  vertices,  the  triangles  formed 
with  the  bases  are  congruent. 

3.  The  diagonals  of  a  square  are  equal. 

4.  If  the  diagonals  of  a  parallelogram  are  equal,  the  figure 
is  a  rectangle. 

5.  Parallels  intercepted  between  parallels  are  equal. 

6.  Prove  the  correctness  of  the  construction  of  Construc- 
tion X,  p.  32. 

7.  Suppose  c  in  Fig.  56  to  be  a  right  angle.     Prove  Za  = 
Z.  b  =  Z  d  =  a  right  angle. 


a 


FIG.  56 


FIG.  57 


8.  /f  the  two  parallels  A  B  and  C  D  are  cut  by  the  trans- 
versal E  F  (Fig.  57),  then: 


J 

,\  Congruency  of  Rectilinear  Figures  and  Circles          41 

(i)   Za=Zc'         (3)   Z6=Zrf'  (5)   Z6'+Zc  =  i8o° 

7(2)   Zc=Za'         (4)   Z&  +  Zc'  =  i8o°  (6)   Zd+Zc'=i8o0. 

g.  Prove,  that  if  any  one  of  the  six  relations  of  Exercise  8 
is  not  true  the  lines  A  B  and  C  D  are  not  parallel. 

10.  Two  of  the  angles  of  a  triangle  have  the  values  indi- 
cated by  the  following  pairs  of  numbers.     Find  the  third  angle 
in  each  case. 

(1)  40°  and  60°          (5)  50  and  a  (9)  $x  and  2X 

(2)  65°  and  65°          (6)  r  and  70  (10)  x  and  y 

(3)  12°  and  98°  .        (7)  A  and  B  (n)  a  and  b+c 

(4)  37|°andi23i°    (8)  A/$  and  B/8      (12)  —  and  2^. 

o  0 

11.  Prove,  if  two  angles  of  one  triangle  are  equal  to  two 
angles  of  another,  the  third  angles  are  equal,  and  the  triangles 
are  mutually  equiangular. 

12.  Prove,  that  an  exterior  angle  of  a  triangle  is  equal  to 
the  sum  of  the  two  opposite  interior  angles  of  the  triangle. 

13.  One  base  angle  of  an  isosceles  triangle  is  J  of  the 
vertex-angle.     Find  the  angles  of  the  triangle. 

14.  Find  the  angle  between  the  bisectors  of  the  acute  angles 
of  a  right  triangle. 

15.  The  bisector  of  an  exterior  vertex-angle  of  an  isosceles 
triangle  is  parallel  to  the  base.     Prove. 

1 6.  How  many  diagonals  can  be  drawn  from  one  vertex 
in  a  quadrilateral?    In  a  pentagon?    In  a  hexagon?    In  a 
heptagon  ?     In  an  octagon  ?    In  an  w-gon  ?    In  a  2«-gon  ? 

1 7.  Into  how  many  triangles  do  the  diagonals  of  Exercise  16 
divide  the   quadrilateral  ?    The  pentagon  ?     The   hexagon  ? 
The  heptagon  ?    The  octagon  ?    The  w-gon  ?    The  2»-gon  ? 

1 8.  Prove,  that  the  sum  of  the  interior  angles  of  an  w-gon 
is  2n— 4  right  angles,  or  (»  —  2)  straight  angles. 


42  Second- Year  Mathematics 

ig.  Prove,  that  any  interior  angle  of  a  regular  w-gon  is 

2n  —  4    .  .  A  (n  —  2\        . 

right  angles,  or  I straight  angles. 

n  \   n   / 

20.  Prove,  that  the  sum  of  the  exterior  angles  of  an  w-gon 
is  4  right  angles. 

The  distance  between  two  parallels  is  measured  on  a  common 
perpendicular  between  them.    See  §  60,  3,  and  §  99. 

21.  Prove,  that  parallels  are  everywhere  equally  distant. 

22.  Prove,  if  two  points  on  the  same  side  of  a  line  are 
equally  distant  from  the  line,  the  line  joining  the  two  points  is 
parallel  to  the  given  line. 

79.  Algebraic  solutions  of  equations  based  on  geometrical 
relations. 

(i7)The  sides  of  an  equiangular  triangle  are  denoted  by 
,  2X—y,  and  14.     Find  x  and  y. 

2.  The  angles  of  an  equiangular  triangle  are  denoted  by 

,  3(3^  —  2^),  and  60.     Find  x  and  y. 

3.  The  three  sides  of  an  equiangular  triangle  are  denoted 
by  >jx+2y,  $(x  +  2y),  and  Sx  —  ^y  +  2.     Find  x  and  y. 

Solve  the  quadratics  in  the  following  by  factoring,  being 
careful  to  obtain  all  solutions. 

4.  One  pair  of  opposite  sides  of  a  parallelogram  is  denoted 
by  x2  -\-x  and  6(3  —x},  and  the  other  pair  by  y2  —y  and  3(5  —y). 
Find  x  and  y  and  the  lengths  of  the  sides. 

5.  Two  opposite  angles  of  a  parallelogram  are  denoted  by 
x2+6  and  7(^+2).     Find  x  and  all  angles  of  the  parallelo- 
gram. 

6.  The  diagonals  of  a  rectangle  are  denoted  by  x2  —x  and 
2(2^+7).     Find  both  values  of  x  and  the  diagonals. 

7.  The  diagonals  of  a  parallelogram  divide  each  other  so 
that  the  segments  of  one  are  s2+s  and  2(55  —  7),  and  of  the 


Congruency  of  Rectilinear  Figures  and  Circles         43 


other,  /2  +  2/  and  8(3—  /).     Find  s,  t,  and  the  lengths  of  the 
diagonals. 

Is  the  parallelogram  a  rectangle  ?     Why  ? 

8.  The  diagonals  of  a  rhombus  divide  each  other  so  that 
the  parts  of  one  diagonal  are  denoted  by  x2  and  3(2^+9), 
and  of  the  other  by  y2  and  2(^+4).     Find  x,  y,  and  both  of 
the  diagonals. 

9.  Two  of  the  four  angles  that  the  diagonals  of  a  rhombus 
make  with  each  other  are  given  by  x2  —  10  and  10(2^—11). 
Find  x  and  all  the  four  angles. 

10.  The  distances  from  a  point  on  the  perpendicular  bisec- 
tor of  a  line-segment  to  the  ends  of  the  segment  are  given  by 
z2  +  3z  and  18(2—3).     Find  z  and  the  distances  from  the  point 
to  the  ends  of  the  line-segment. 

11.  The   two   sides  that   include  the  vertex-angle  of  an 
isosceles  triangle  are  given  by  x2—  x  and  6(x  —  2).     Find  x 
and  the  lengths  of  the  sides. 

12.  The  angles  opposite  the  equal  sides  of  an  isosceles 
triangle  are  given  by  x3  —  i   and  3(^+9).     Find  x  and  the 
lengths  of  the  sides. 

13.  Two  angles  of  a  triangle  are  equal.     The  sides  oppo- 
site these  two  angles  are  given  by  x2  +  $  and  3(^  +  5).     Find 
x  and  the  lengths  of  the  sides. 

Sides  and  angles  of  polygons  may  be  denoted  by  either  positive  or 
negative  numbers,  since  the  sign  of  a  number  only  indicates  the  direction 
in  which  the  line  or  angle  is  conceived  as  measured. 

14.  The  angles  opposite  the  equal  sides  of  an  isosceles  tri- 
angle are  given  by  xa  —  i  and  8(#  +  i).     Find  x  and  the  values 
of  the  angles. 

15.  The  lengths  of  the  bisectors  of  the  base  angles  of  an 
isosceles  triangle  are  given  by  m3  and  i6(w—  4).     Find  m  and 
the  lengths  of  the  bisectors. 


44  Second-Year  Mathematics 

1 6.  The  lengths  of  the  perpendiculars  from  the  vertices  of 
the  base  angles  of  an  isosceles  triangle  to  the  opposite  sides 
are  given  by  r2+4r  and  12(2^—3).     Find  r  and  the  lengths 
of  the  perpendiculars. 

17.  The  lengths  of  the  two  lines  drawn  from  the  vertices 
of  the  base  angles  of  an  isosceles  triangle  to  points  of  the 
opposite  sides  equidistant  from  the  vertex  of  the  vertex-angle 
are  given  by  c(c  +  $)  and  21  (c— 3).     Find  c  and  the  lengths  of 
the  two  lines. 

1 8.  The  following  pairs  of  numbers  denote  the  lengths  of 
the  sides  lying  opposite  a  pair  of  equal  angles  of  a  triangle. 
Find  x  and  the  lengths  of  the  sides  in  each  case. 

(1)  x2  and  5#— 4  (10)  3(5^  +  1)  and  14* 

(2)  x2  +  i  and  2(#+i8)  (n)  4(y^2  — i)  and  gx 

(3)  x2  —  2  and  3#— 4  (12)  $x2  and  2  —  5^ 

(4)  x2—  4  and  x  +  i6  (13)  $x2  and  57—103: 

(5)  x2~4  and  4(2—  x)  (14)  4X2  and  3(^  +  9) 

(6)  x2  —  i  and  2(^  +  7)  (15)  4gx2  —  i  and  5(13  — 73:) 

(7)  x2  +x  and  4(9—  x)  (16)  iSx2  and  3(^  +  12) 

(8)  x(x  +  i)  and  2—$x*  (17)  40  and  ^x2  +  ijx 

(9)  Sx2  and  2^+3  (18)  x*—gx*  and  2(12—133;)* 
The  only  difficult  step  in  solving  the  equation  of  Example 

(18)   is  in  factoring  the   first   member.     We  apply  here  the 
Remainder  or  Factor  Theorem  (see  FYM,  p.  314,  §261). 

Let  us  rehearse  the  reasoning  given  in  the  reference.  Fac- 
toring is  finding  exact  divisors.  Exact  divisors  are  divisors 
that  give  o  for  the  remainder  after  dividing.  The  quickest 
way  to  discover  exact  divisors,  or  factors,  is  to  examine  the 
remainders.  But  the  Remainder  Theorem  states  that  the  result 
of  substituting  any  number  (say  8  or  r)  for  x  in  an  expression 

*  Example  (18)  leads  to  the  equation  xi— gx2  +  26x—  24  =  0.  In 
factored  form  this  is  (x—  z)(x—  $)(x—  4)  =o.  The  values  of  x  are 
then  x  =  2,  #  =  3,  and  00  —  4. 


Congruency  of  Rectilinear  Figures  and  Circles         45 


containing  x,  is  the  same  as  the  remainder  after  dividing  the 
expression  by  x— 8,  or  x— r.  Hence,  to  find  factors  we  begin 
with  o  and  substitute  i,  2,  3,  etc.,  then  — i,  —  2,  —3,  etc.,  until 
the  expression  we  are  trying  to  factor  gives  a  result  o. 

x  minus  the  number  that  substituted  for  x  gives  o,  is  then  a 
factor  of  the  expression. 

The  graph  of 
x3—  <)x2+26x  —  24  (Fig. 
58),  shows  that  there 
are  3,  and  only  3,  values 
of  x  that  will  make  the 
expression  equal  o. 
These  are  the  three  dis- 
tances from  o  to  the 
crossing-points  of  the 
curve  on  the  horizontal 
axis.  There  are  no  other 
crossing-points  than 
those  shown  because 
above  #=4  and  below 
x  =  i  the  curve  leaves 
the  horizontal  axis 
farther  and  farther. 

Similarly  the  graph 
shows  that  there  are 
as  many  crossing-points, 


FIG.  58 


and  hence  real  values  of  x  where  the  expression  equals  o, 
as  there  are  units  in  the  highest  exponent  of  x  in  the  expres- 
sion plotted. 

19.  Solve  the  following  equations  by  this  method  of  fac- 
toring: 

(1)  X3—&X3  +  I()X  — 12=0  (4)    #3—  4X*+X  +  6=0 

(2)  X3—  lJXa  +  I4X-?S=0  (5)    X3—X'  —  IOX  —  S=0 

(3)  X3~ 3*a—  #+3=0  (6)  x^  —  T> 


46 


Second-  Year  Mathematics 


(7)  x4  —  7 

(8)  x4-$ 

(9)  X4  —  I 

(lo)  X4-\-2X3—I^X 


—  8x=o 


14^  +  24=0. 


___  __  CIRCLES 

80.  The  curve  which  bounds  a  circle  is  the  circumference. 

8  1.  A  line-segment  drawn  through  the  center  and  termi- 
nated both  ways  by  the  circumference  is  a  diameter,  as  B  C 
(Fig.  59)- 

82.  A  line-segment  from  the  center  to  the  circumference 


Tariff  e-nt 


is  the  radius,  as  O  A. 

83.  Any     line-segment 
terminated  both  ways  by 
the    circumference     is     a 
chord. 

84.  A  line  cutting  the 
circumference  in  two  points 
is  a  secant. 

85.  A  line  that  touches 
the  circumference   in  but 

one  point,  however  far  the  line  is  prolonged,  is  a  tangent. 

86.  Any  part  of  the  circumference  is  an  arc,  as  A  C  (Fig.  59). 

87.  The  parts  into  which  a  chord  or  a  secant  divides  the 
circle,  are  circular  segments,  or  simply  segments. 

88.  The  portion  of  the  circle  included  between  two  radii 
and  the  arc  they  intercept,  is  a  sector. 

Circles  are  designated  by  reading  the  letter  at  their  centers, 
as  the  circle  O  (Fig.  59). 


FIG.  59 


CONSTRUCTION  XI 
89.  Draw  two  circles  having  equal  radii. 


Congruency  of  Rectilinear  Figures  and  Circles 
PROPOSITION  XXII 


47 


90.  Theorem:     Tico  circles  having  equal  radii  are  equal, 
and  conversely,  equal  circles  have  equaljradj4._ 

Given  the  circles  whose  centers  are  O  and  O',  and  having 
A  =  O'A'  (Fig.  60). 


FIG.  60 

To  prove  the  circles  are  equal. 

Proof:  Place  the  circle  O  upon  the  circle  O'  so  that  the 
center  O  falls  on  the  center  O'. 

The  circles  then  coincide  throughout,  for  if  they  did  not, 
there  would  be  points  of  one  circumference  farther  from  O 
and  O'  than  some  points  of  the  other,  and  for  such  points 
the  distances  from  O  and  O'  would  not  be  equal. 

Conversely,  since  the  circles  are  equal  they  may  be  made 
to  coincide,  and  hence  the  radii  are  equal. 

EXERCISE 
i.  The  diameters  of  equal  circles  are  equal. 

PROPOSITION  XXIII 

91.  Problem:  Given  the  circumference  of  a  circle,  to  find 
the  center. 


Second-  Year  Mathematics 


I.     Draw  any  two  non-parallel  chords  in  the  circle  (see 


DF 


Fig.  61). 

Bisect  each  of  the  chords  with 
perpendiculars. 

These  perpendiculars  will  inter- 
sect (why  ?)  at  some  point,  as  at  O. 

O  is  the  required  center. 

Assume  that  the  perpendicular  bi- 
sector of  a  line-segment  contains  every 
point  that  is  equally  distant  from  the 
end-points  of  the  line-segment.  (Proved 
on  p.  174.) 

II.  Draw  a  single  chord  as  AB,  and  bisect  it  as  with 
E  F  _L  to  it. 

The  middle  of  E  F,  as  O,  is  the  center  of  the  circle.     Why  ? 

PROPOSITION  XXIV 

92.  Theorem:    A  diameter  divides  the  circumference  of  a 
circle  into  two  equal  parts;  and  conversely,  any  chord  that  divides 
a  circumference  into  two  equal  parts,  is  a  diameter. 

I.  Draw  a  circle  with  a  diameter,  and  show  that  if  the  cir- 
cle be  folded  over  this  diameter  as  an  axis,  or  hinge,  the  one 
part  will  coincide  with  the  other. 

II.  The  proof  of  the  converse  is  left  to  the  student. 

PROPOSITION  XXV 

93.  Theorem:    In  the  same  circle  or  in  equal  circles,  radii 
that  form  equal  angles  intercept  equal  arcs  on  the  circumference, 
and  conversely. 

I.  Given  in  circle  O,  the  angle  A  O  B  made  by  the  radii 
O  A  and  O  B,  equal  to  angle  COD  made  by  the  radii  .O  C 
and  O  D  (Fig.  62). 

To  prove  arc  A  B  =arc  CD. 

Proof :   Imagine  the  plane  of  the  circle  cut  along  A  O  and 


Congruency  of  Rectilinear  Figures  and  Circles         49 

B  O  and  the  sector  A  O  B  turned  round  O  as  a  pivot,  until 
O  A  falls  on  O  C. 

Then  A  will  fall  on  C.     Why  ? 

O  B  will  fall  along,  or  will  take  the  direction  of,  O  D. 
Why? 

B  will  fall  on  D.     Why  ? 

j 


FIG.  62 

Arc  A  B  will  coincide  with  arc  CD,  or  the  points  of  the 
given  circumference  would  be  unequally  distant  from  O. 
Hence  arc  A  B=arc  C  D. 

II.  Given  circle  O=circle  O',  and  angle  A  O  B,  made  by 
the  radii  O  A  and  O  B,  equal  to  angle  A'O'B',  made  by  the 
radii  O'A'  and  O'B'. 

To  prove  arc  A  B=arc  A'B'. 

Proof:  Place  the  circle  O  upon  the  circle  O'  so  that  O  falls 
on  O'  and  O  A  along  O'A'. 

Then  A  will  fall  on  A7.     Why  ? 

O  B  will  take  the  direction  O'B'  (why  ?)  and  B  will  fall 
onB'.  Why? 

Hence,  arc  A  B  =  arc  A'B'. 

III.  Converse.     Given  circles   O   and  O'  equal  and  arc 
AB  =  arc  CD  =  arc  A'B'. 

To  prove  ZA  O  B  =  ZC  O  D  =  Z  A'O'B'. 


5o  Second-Year  Mathematics 

Proof:  Suppose  the  arc  A  B  to  be  turned  around  O  as 
a  pivot  until  arc  A  B  coincides  with  C  D. 

Prove  that  the  sectors  A  O  B  and  COD  coincide. 

Next  place  circle  O  upon  circle  O'  so  that  arc  A  B  coincides 
with  arc  A'B'.  Why  can  this  be  done  ? 

Then  prove  sector  A  O  B  to  coincide  with  sector  A'O'B'. 

Hence,  in  both  cases  the  angles  at  the  centers  are  equal. 

94.  Angles  whose  vertices  lie  at  the  center  of  a  circle  are 
called  central  angles. 

EXERCISE 

i.  Prove,  in  the  same  or  in  equal  circles  radii  that  form 
unequal  angles  intercept  unequal  arcs  ori  the  circumference, 
and  conversely. 

PROPOSITION  XXVI 

95.  Theorem:    In  the  same  circle  or  in  equal  circles,  equal 
arcs  are  subtended  by  equal  chords,  and  conversely. 

I.     Given  circle  O,  with  arc  A  B=arc  CD  (Fig.  63). 

fi 


C 

FIG.  63 

To  prove  chord  A  B=chord  C  D. 
This  is  left  as  an  exercise  for  the  student. 

II.  Given  circle  O=circle  O',  and  arc  A  B=arc  A'B'. 
To  prove  chord  A  B  =chord  A'B'. 

III.  Conversely,  given  circle  O,  with  chord  AB=chord 
CD. 


Congruency  of  Rectilinear  Figures  and  Circles          51 


To  prove  arc  A  B  =  arc  C  D. 

Join  the  ends  of  the  equal  arcs  with  the  center,  or  centers,  prove 
the  triangles  thus  formed  congruent,  the  central  angles  equal,  and  apply 
Proposition  XXV. 

Let  the  student  arrange  the  proof. 

EXERCISES 

1.  If  a  circumference  is  divided  into  three  equal  parts, 
and  the  points  of  division  are  joined  by  chords,  the  polygon 
formed  is  an  equilateral  triangle.     Prove. 

2.  The  ends  of  a  pair  of  perpendicular  diameters  are  joined. 
What  kind  of  a  polygon  is  formed  by  the  joining  lines  ?   Prove 
your  answer. 

3.  Draw  a  circle  O=  circle  O',  and  a  chord  A  B=  chord 
A'B'.     Prove  arc  A  B  =arc  A'B'. 

PROPOSITION  XXVII 

96.  Theorem:  A  radius,  drawn  perpendicular  to  a  chord, 
bisects  the  chord  and  also  the  arc  subtended  by  the  chord. 

Given  circle  O  and  radius  O  C  drawn  perpendicular  to 
the  chord  A  B  (Fig.  64),  intersecting 
the  chord  at  D. 

To  prove  that  chord  A  B  and  arc 
A  B  are  bisected  by  O  C. 

Join  O  A  and  O  B.     What  kind  of 
triangle  is  A  O  B  ? 

Prove  triangles  A  O  D  and  BOD 
congruent. 

Then  apply  Proposition  XXV,  to 
prove  arc  A  C=B  C. 

EXERCISES 

1.  The  diameter  E  C_L  A  B  bisects  the  arc  A  E  B.    Prove. 

2.  A  diameter  that  bisects  a  chord  is  perpendicular  to  the 
chord,  and  bisects  the  subtended  arc.     Prove. 


52  Second- Year  Mathematics 

3.  A  line  bisecting  a  chord  and  the  subtended  arc  passes 
through  the  center,  and  is  perpendicular  to  the  chord.     Prove. 

4.  Show  how  to  bisect  a  given  arc. 

5.  Through  a  given  point  within  a  circle  draw  a  chord  that 
will  be  bisected  by  the  point. 

6.  Prove  the  line-segment  connecting  the  middle  point  of 
the  two  arcs  into  which  a  chord  divides  a  circumference,  is  a 
diameter  perpendicular  to  the  chord. 

7.  Prove  a  diameter  bisecting  an  arc  is  the  perpendicular 
bisector  of  the  chord  that  subtends  the  arc. 

8.  The  perpendicular  bisector  of  a  chord  passes  through 
the  center  of  the  circle  and  bisects  the  subtended  arc.     Prove. 

9.  Draw  a  circle  through  two  fixed  points.     How  many 
such  circles  may  be  drawn  ? 

10.  Draw  a  circle  passing  through  three  fixed  points  not 
situated  on  a  straight  line.     How  many  such  circles  may  be 
drawn  ? 

n.  A  line  perpendicular  to  a  chord  and  bisecting  the 
subtended  arc  passes  through  the  center  of  the  circle  and  bisects 
the  chord.  Prove. 

12.  From  Proposition  XXVII  and  Exercises  2,  3,  7,  8,  and 
ii  write  down  four  conditions  for  a  line  to  fulfil  about  the 
center  of  a  circle,  a  chord  and  the  subtended  arc,  and  show 
that  if  a  line  fulfils  any  two  of  the  conditions,  the  remaining 
two  are  also  fulfilled. 

PROPOSITION  XXVIII 

97.  Theorem:  In  the  same  circle,  or  in  equal  circles,  equal 
chords  are  equally  distant  from  the  center,  and  conversely. 

Assume  that  the  distance  from  a  point  to  a  line  is  measured 
on  the  perpendicular  from  the  point  to  the  line.  (Proved  in  §  99.) 

Given  in  the  circle  O,  or  in  the  circles  O  and  O',  chord  A  B 
equal  to  chord  A'B'  (Fig.  65). 


Congruence  of  Rectilinear  Figures  and  Circles          53 


To  prove  O  C  =  O  C',  and  O  C  =  O'C'. 

Draw  O  A,  O  A',  and  O'A'. 

Prove  triangles  O  A  C,  O  A'C',  and  O'A'C'  congruent. 

Conversely,  given  O  C  =  O  C'  =  O'C'. 


FIG.  65 

To  prove  chord  A  B=chord  A'B'  in  either  circle. 
Prove  triangles  O  A  C,  O  A'C',  and  O'A'C'  congruent. 

98.  By  the  distance  from  a  point  to  a  line  is  meant  the 
shortest  distance. 

PROPOSITION  XXIX 

99.  Theorem:    The  distance  from  a  point  to  a  line  is  meas- 
ured on  a  perpendicular  from  the  point  to  the  line. 

GivenPD_LAB  (Fig.  66). 

To  prove  P  D  shorter  than  any 
other  line  drawn  from  P  to  A  B. 

Let  P  E  be  any  other  line  than  P  D 
from  P  to  A  B. 

Prolong  P  D  to  P',  making 
P/D=PD. 

Join  E  and  P'.  EP'=EP.  Why? 

P  E  P'  is  not  a  straight  line.  Why  ? 

P  P'  is  then  shorter  than  P  E  +E  P'.    Why  ? 


54 


Second-  Year  Mathematics 


Hence,  half  of  P  P',  or  P  D,  is  shorter  than  half  of  P  E  + 
EP'.  Why? 

P  E  being  any  other  line  than  the  perpendicular  P  D,  the 
line  P  D  is  the  shortest  from  P  to  A  B. 

NOTE. — See  proof  of  the  converse  of  this  proposition  in  the  chapter 
on  Inequalities,  §  208.  State  the  converse. 

100.  The  point  where  a  tangent  to  a  circle  touches  the 
circumference,  is  called  the  point  of  tangency,  or  the  point  of 
contact. 

PROPOSITION  XXX 

101.  Theorem:    A  tangent  to  a  circle  is  perpendicular  to 
the  radius  drawn  to  the  point  of  contact;  and  conversely,  a  line 
drawn  perpendicular  to  a  radius  at  its  outer  end  is  a  tangent  to 
the  circle. 

I.     Let  A  B  be  tangent  to  the  circle  at  the  point  C  (Fig.  67). 

To  prove  the  radius  O  C_LA  B. 

Let  D  be  any  other  point  of 
A  B  than  C. 

Join  D  with  O.  Then  O  D  is 
longer  than  O  C.  Why  ? 

Hence,  all  lines  drawn  from 
O  to  other  points  of  A  B  than  C 
are  longer  than  O  C.  Why  ? 

Therefore,  O  C_LA  B.  Why? 
The  truth  of  the  converse  is  here 


D 


FIG.  67 


See  note  under  §  99. 
assumed. 

II.     Conversely,  let  A  B  be  _LC  O,  a  radius,  at  C. 

To  prove  A  B  tangent  to  the  circle. 

C   is    a   point   common   to   the   line   A  B    and   the   cir- 
cumference.   Why  ? 

Let  the  student  prove  any  other  point  of  A  B,  as  D,  farther 


Congruency  of  Rectilinear  Figures  and  Circles          55 

from  O  than  the  circumference,  and  hence  all  points  of  A  B 
other  than  C  to  be  outside  the  circle. 

Therefore  A  B  is  tangent  to  the  circle  at  C. 

PROPOSITION  XXXI 

1 02.  Theorem:     The  arcs  included  between  two  parallel 
secants  are  equal;   and  conversely,  if  two  secants  include  equal 
arcs,  and  do  not  intersect  within  the  circle,  they  are  parallel. 

I.  Given  circle  O  and  A  B  ||  C  D,  both  cutting  the  circum- 
ference (Fig.  68)  at  these  points. 

To  prove  arcs  A  C  and  B  D  equal. 

Draw  OE  through  the  center  _L  AB, 
and  prolong  it. 

This  line  is  also  _L  C  D  (§  60, 
Exercise  3,  p.  30). 

A~E = E~B*  and  E~C  =  E~D .    Why  ? 

Complete  the  proof. 

II.  Conversely,    given   A  C=B  D 

(Fig.   68),   A  B   and   C  D   not   intersecting    within   the   cir- 
cumference. 

To  prove  A  B  ||CD. 

Draw  O  E  _L  A  B  and  prolong  it  to  F. 

Prove  E~C=ED  and  CfF=D>. 

Then  refer  to  p.  52,  Exercise  6,  and  complete  the  proof. 

III.  Prove  the  theorem  with  one  of  the  lines,  as  A  B, 
tangent  to  the  circle,  as  in  Fig.  69. 

IV.  Prove  the  theorem  with  both  parallels  tangent  to  the 
circle,  as  in  Fig.  70. 

103.  Two  circles  are  said  to  be  tangent  to  each  other  if 
both  are  tangent  to  the  same  line  at  the  same  point.     This 

*  A  E  and  E  B  mean  are  A  E  and  are  E  B. 


Second-  Year  Mathematics 


point  is  the  point  of  tangency,  or  the  point  of  contact  of  the 
circles. 


If  the  tangent  circles  lie  wholly  without  each  other  they 
are  tangent  externally  (Fig.  71). 

If  one  of  the  tangent  circles  lies  within  the  other  they  are 
tangent  internally  (Fig.  72). 

PROPOSITION  XXXII 

104.  Theorem:  If  two  circles  are  tangent  to  each  other, 
their  centers  and  the  point  of  tangency  lie  in  a  straight  line. 

I.  Let  O  and  O'  be  the  centers  of  two  circles  tangent 
externally,  T  being  the  point  of  tangency  (Fig.  71). 


FIG.  71  FIG.  72 

To  prove  O,  O',  and  T  lie  in  a  straight  line. 


Congruency  of  Rectilinear  Figures  and  Circles          57 

Prove  that  O  T  O'  is  a  straight  angle. 

Prove  that  O  T  and  O'T  are  in  a  straight  line,  by  Proposition  VII, 
p.  22. 

II.     Prove  the  proposition  for  the  case  shown  in  Fig.  72. 
EXERCISES 

1.  If  two  circles  are  tangent  to  each  other  externally,  the 
distance  between  their  centers  is  the  sum  of  their  radii.     Prove. 

2.  If  the  distance  between  the  centers  of  two  circles  is  equal 
to  the  sum  of  their  radii  the  circles  are  tangent  externally.  Prove. 

3.  If  two  circles  are  tangent  internally  the  distance  between 
their  centers  is  equal  to  the  difference  between  the  radii.  Prove. 

4.  Draw  a  circle  of  given  radius  tangent  to  a  given  circle 
(a)  externally,  (b)  internally. 

How  many  such  circles  can  be  drawn  tangent  to  the  same 
circle  ? 

In  what  line  do  the  centers  lie  ? 

5.  Draw  a  circle  tangent  to  a  given  circle  at  a  given  point 
in  the  circumference. 

How  many  such  circles  can  be  drawn  ? 
In  what  line  do  the  centers  lie  ? 

6.  Draw  a  circle  through  a  given  point  and  tangent  to  a 
given  circle. 

7.  The  distance  between  the  centers  of  two  tangent  circles 
is  2^  inches.     The  radius  of  one  is  f  inch.     Draw  the  two 
circles. 

8.  Draw  a  common  tangent  to  two  tangent  circles. 

9.  How  many  circles  can  be  drawn  through  the  same  two 
points  ? 

In  what  line  do  the  centers  lie  ? 

10.  The  radii  of  three  circles  are  i  inch,  ij  inches,  and 
f  inch,  respectively.     Draw  the  circles  tangent  to  each  other 
externally. 


58  Second-Year  Mathem-atics 

PROPOSITION  XXXIII 

105.  Theorem:    The  line  joining  the  centers  of  two  inter- 
secting circles  bisects  the  common  chord  perpendicularly. 

Let  O  and  O'  be  the  intersecting  circles  (Fig.  73). 


The  student 


FIG.  73 

To  prove  O  O'_LA  B. 

For  proof  see  Proposition  XI,  p.  25,  for  both  positions, 
will  supply  the  proof. 

EXERCISES 

1.  ABC  is  a  right  triangle,  /_A  being  an  acute  angle. 
Let  o  denote  the  length  of  the  side  opposite  Z.A,  a,  the  length 
of  the  side  adjacent  to  Z.A,  and  h,  the  hypotenuse.     Express 
in  these  numbers  the  following  ratios: 

(1)  the  opposite  side  to  the  hypotenuse 

(2)  the  adjacent  side  to  the  hypotenuse 

(3)  the  opposite  side  to  the  adjacent  side 

(4)  the  hypotenuse  to  the  opposite  side 

(5)  the  hypotenuse  to  the  adjacent  side 

(6)  the  adjacent  side  to  the  opposite  side. 

2.  Denote  these  six  ratios  as  follows: 

s—o/h  C=h/o 

c=a/h  S=h/a 

t  —  o/a  T=a/o 

and  express  C  in  terms  of  s;  S  in  terms  of  c;   T  in  terms  of  t. 


Congruency  of  Rectilinear  Figures  and  Circles          59 


3.  It   has   been   shown    (see   FYM,   pp.    198,    199).    that 
o*+a2=h2.     Divide  both  sides  of  o2+a2=h2  by  h2 .  and  ex- 
press the  resulting  equation  in  terms  of  .s  and  c. 

4.  Solve  the  equation,  s2+c2  =  i,  for  s;  for  c. 

5.  Express  C  in  terms  of  c;  5  in  terms  of  s;    T  in  terms 
of  5  and  c;   T  in  terms  of  s;   T  in  terms  of  c. 

6.  Express  c,  t,  C,  S,  and  T  in  terms  of  s. 

7.  Express  5,  t,  C,  S,  and  T  in  terms  of  c. 

8.  Express  s,  c,  C,  S,  and  T  in  terms  of  /. 

9.  Express  s,  c,  t,  S,  and  T  in  terms  of  C. 

10.  Express  s,  c,  t,  C,  and  T  in  terms  of  S. 

1 1 .  Express  s,  c,  t,  C,  and  5  in  terms  of  T. 


a 


FIG.  74 

12.  In  a  right  triangle  0  =  10,  and  s  =  %.     Find  h,  C,  c,  a, 
S,  t,  and  T. 

13.  In  a  right  triangle  a =8  and  c  =  J.     Find  h,  S,  s,  o, 
C,  t,  and  T. 

14.  In  a  right  triangle  ^  =  15  and  s=^.     Find  0,  C,  c, 
S,  a,  t,  and  T. 

15.  Two  circles  described  on  the  diagonals  of  a  rectangle 
have  areas  denoted  by  x2—  2X  and  4(7^  —  26).     Find  x  and 
the  areas  of  the  circles. 

1 6.  Describe  a  circle  that  shall  pass  through  two  given 
points  and  have  a  given  radius. 


6o 


Second-  Year  Mathematics 


17.  A  diameter  that  bisects  a  chord  bisects  the  central 
angle  between  radii  drawn  to  the  ends  of  the  chord.     Prove. 

1  8.  The  perpendicular  bisectors  of  the  sides  of  an  inscribed 
quadrilateral  meet  at  a  common  point.     Prove. 

19.  If  two  intersecting  chords  make  equal  angles  with  a 
line  joining  their  common  point  to  the  center,  the  chords  are 
equal.     Prove. 

20.  Prove  Za  (Fig.  75)  is  two  times  Z#.     The  vertex  of 
a  is  the  center  of  the  circle. 


FIG.  75 


FIG.  76 


21.  Draw  a  pan*  of  perpendicular  diameters  in  a  circle, 
connect  their  consecutive  ends  and  prove  the  figure  formed 
by  the  connecting  lines  a  square. 

22.  Draw  a  circle,  step  around  it  using  the  radius  for  a 
step,  connect  every  alternate  mark,  and  prove  that  the  con- 
necting lines  form  an  equilateral  triangle. 

23.  A  B  C  is  an  equilateral  triangle  and  C  D  is  perpen- 
dicular to  A  B   from  the  vertex  C  (Fig.  76).     Express   the 
six  ratios  s,  c,  t,  C,  S,  and  T  in  triangle  ADC  (see  Problem 
2,  p.  58). 

24.  Express  a  and  o  in  terms  of  h  (see  Problem  3,  p.  58). 

25.  Calculate  the  numerical  values  of  all  six  of  the  ratios, 
s,  c,  t,  C,  S,  and  T  for  angle  A,  using  triangle  ADC  (Fig.  76). 


Congruency  of  Rectilinear  Figures  and  Circles         61 


26.  What  is  the  numerical  value  in  degrees  of  Z.  A  ?    Of 
?    Of  one  of  the  parts  of  ZC  made  by  C  D  ? 

27.  Denoting  the  ratios  for  a  60°  angle  by  s^o,  C6o>  *6o> 
•$60,  Ceo,  and  Tt0,  calculate  the  six  numerical  values  from 
triangle  A  D  C  of  Fig.  76. 

28.  Denoting  the  six  ratios  for  a  30°  angle  similarly,  cal- 
culate the  numerical  values  of  s30,  c30,  t30,  C30,  S30,  and  T30, 
using  triangle  ADC  (Fig.  76).  ^ 

29.  ABC  (Fig.  77)  is  a  right  triangle 
having     Z.A  =  /.B=^°.       Compare     a 
and  o.     Using  a2+02=/i2  express   h   in 
terms  of  o;  of  a;  o  in  terms  of  h;  of  a; 

a  in  terms  of  h.  /}        Q.         C 

•  30.  Calculate    from    triangle    ABC  FlG'  77 

(Fig.  77)  the  numerical  values  of  s4S,  c4S,  t4S,  C4S,  S4S.  and  T4S. 

31.  Given  540=.643.  Calculate  to  three  decimal  places 
the  numerical  values  of  c40,  /40,  C40,  S40,  T40,  using  triangle 
ABC  (Fig.  78). 

Use  §  105,  Exercise  4. 


451 


FIG.  78 


FIG.  79 


32.  Given  sso=.766.      Calculate  to  three  decimal  places 
the  values  of  cso,  tso,  Cso,  S50,  and  Tso,  using  triangle  ABC 
(Fig.  78). 

33.  In  a  regular  hexagon  (Fig.  79),  given  the  side  I  =  12 
and  530=^.    Calculate  the  radius  of  the  inscribed  circle. 


62  Second-Year  Mathematics 

34.  In  a  regular  octagon  (Fig.  80),  given  the  side  I  equal 
to  1 6  and  c32^=  .924.  Calculate  the  radius  of  the  inscribed 
circle. 


FIG.  80 


35.  In  a  regular  nonagon,  given  the  side  £  =  24  and 
.  364.     Find  the  radius  of  the  inscribed  circle. 


CHAPTER  II 

RATIO,   PROPORTION,    SIMILAR   TRIANGLES 
Ratio  of  Numbers  and  Segments 

1 06.  To  measure  the  number  a  by  the  number  b,  a  is 
divided  by  b.     The  quotient  a/b  is  the  ratio  of  a  to  b.     Thus, 
the  ratio  of  a  to  b  indicates  how  many  times  b  is  contained  in 
a,  or  what  fractional  part  a  is  of  b. 

The  numbers  a  and  b  are  the  terms  of  the  ratio  a  to  b; 
the  first  number  a  is  the  antecedent,  and  b  the  consequent. 

107.  The  portion  of  a  line  bounded  by  two  points  on  the 
line  is  a  line- segment,  or  briefly,  a  segment. 

1 08.  To  measure  a  segment  m,  is  to  lay  off  on  m  repeatedly 
a  smaller  segment,  as  n  (see  Fig.  81). 

109.  The  number  of  times  a  segment  b  is  contained,  without 
remainder,  in  a  segment  a  is  called  the  numerical  measure  of 
a  in  the  unit  b.     b  is  called  a  unit  of  measure. 

Thus  the  numerical  measure  of  A  B  (Fig.  81)  is  4.     CD  is 
a  unit  of  measure  of  A  B. 

\(.m)     A- • • • 'B 

(n)      0 -D 

FIG.  81 


no.  The  ratio  of  two  segments  is  the  ratio  of  the  numerical 
measures  of  the  segments,  both  segments  being  measured  with 
the  greatest  common  unit. 

But  it  can  be  shown  that  if  any  other  common  unit  of 
measure  be  taken,  it  will  be  contained  in  the  greatest  common 
unit  an  integral  number  of  times.  Say  it  is  contained  m  times. 

63 


64  Second-Year  Mathematics 

Then  both  numbers  expressing  the  measures  of  the  magni- 
tudes will  be  m  times  as  large  as  before,  but  the  ratio  will  be 

the  same,  e.  g.,  -  = — '-^=2 
'7     m.'j     7 

Ratio  of  Other  Magnitudes 

ill.  The  ratio  of  two  angles  or  of  two  arcs  or  areas  or  of 
any  other  two  magnitudes  of  the  same  kind  is,  as  in  the  case 
of  segments,  the  ratio  of  the  numerical  measures,  both  magni- 
tudes being  measured  with  the  same  unit. 

Hence,  the  ratio  of  magnitudes  is  a  number.  It  is  the  ratio 
of  the  numerical  measures  of  the  magnitudes  which  are  to  be 
compared.  Consequently  the  ratio  of  two  arcs  may  be  the 
same  as  (equal  to)  the  ratio  of  two  angles  or  the  ratio  of  two 
segments,  etc. 

EXERCISES 

i.  In  Fig.  82,  what  is  the  numerical  measure  of  A  B  ?   Of 
/  .  p  CD?    What  is  the 


FIG.  82 


ratio  of  A  B  to  C  D  ? 
2.  What    is    the 
ratio  of  M  N  to  P  Q 
(Fig.    83)?     What     is     the 
numerical  measure  of  M  N  ? 
Of  PQ? 

3.  In    Fig.    84,   what    is 
the  ratio  of  htoh'?    Of  R  to  R' ? 


o 
p. 

M- 


FIG.  83 


7? 


FIG.  84 


Ratio,  Proportion,  Similar  Triangles 


4.  Draw,   on  squared  paper,   a  figure  like   Fig.  85  with 
D  E  ||  B  C. 

Measure  AD,  D  B,  A  E,  and 
E  C  correctly  to  two  decimal  places. 

Find  the  ratio  of  AD  to  D  B, 
of  A  E  to  EC.  How  do  these 
ratios  compare? 

Squared  paper  may  be  used  in  getting 
the  lengths  of  the  segments. 

5.  Draw  a  figure   like   Fig.  85. 

Using  i  centimeter  as  the  unit,  measure  A  D  and  D  B  and 
find  the  ratio  AD:DB.  Compare  this  ratio  with  the  ratio 
of  A  E  to  E  C,  found  by  using  i  inch  as  the  unit  of  measure. 

6.  What  is  the  ratio  of  Z  A  B  C  to  Z  A'B'C'  (Fig.  86)  ? 


FIG.  85 


FIG.  86 

7.  Cut  from  cardboard  a  circle.  By  rolling  this  circle 
along  a  straight  line,  determine  the  length  of  the  circle  (cir- 
cumference). What  is  the  ratio  of  the  circumference  to  the 

length  of  the  diameter? 

8.  Determine  the 
ratio  of  the  perimeter 
of  a  polygon  to  one  of 


m- 


~b/_ 


'XT 


FIG.  87 


its  sides,  by  using  a 
method  like  that  in 
Problem  7. 


66  •     Second-Year  Mathematics 

9.  Draw  a  figure  like  Fig.  87,  making  lines  I,  II,  III,  etc., 
parallel.     Compare  a/a',  b/b',  c/cf  (use  squared  paper). 

State  the  result  as  a  general  theorem. 

10.  Draw  a  triangle,  as  A  B  C  (Fig.  88).     Draw  the  bisector 

AD  A  B 

of  angle  B.     Compare  the  ratio  — -  with  =--^  (use  squared 

^  _L/       V-'  -D       \^r 

paper). 


C 

FIG.  88 

Commensurable  and  Incommensurable  Segments  and 

Magnitudes 

112.  In  the  preceding  problems,  the  ratio  of  two  magni- 
tudes was  found  by  measuring  each  and  finding  the  ratio  of 
the  numerical  measures. 

A  common  measure  of  two  segments  can  be  found  as  follows: 
Let  A  B  and  C  D  be  the  segments  to  be  measured  (Fig.  89). 
Lay  off  the  smaller  segment  C  D  on  the  larger  segment  A  B, 
leaving  a  remainder,  as  E  B,  which  is  less  than  C  D. 

A  £       & 


c  r  D 

FIG.  89 

.  Lay  off  EB  on  CD  leaving  a  remainder  FD  (<EB). 
Lay  off  F  D  on  E  B.  Suppose  there  is  now  no  remainder. 
Then  the  last  remainder  F  D  is  a  common  measure  of  A  B 
and  C  D. 


Ratio,  Proportion,  Similar  Triangles  67 

For  AB  =  3CD+EB=3(CF+FD)+EB.     (Why?) 
=3(2EB  +  FD)+3FD.     (Why?) 
=3(2.3FD+FD)+3FD.     (Why?) 
=  3(7FD)+3FD  =  24FD.     (Why?) 

Similarly,  CD  =  2EB+FD.     (Why ?) 

=  2.3FD  +  FD  =  7FD.     (Why?) 

Thus  F  D  is  a  common  measure  of  A  B  and  C  D. 
This  process  is  based  upon  the  following  law: 
If  two  magnitudes,  a  and  al}  have  a  common  measure,  let 
al  the  smaller,  be  used  as  a  measure  of  the  larger  a.     If  there 
is  a  remainder,  say  a2,  this  is  less  than  al.     Using  a2  as  a 
measure  of  oI;  the  former  measure,  if  there  is  a  remainder  as 
a3,  then  a3  is  less  than  a2.     If  this  process  is  continued,  a 
remainder,  an,  will  be  reached  which  is  contained  exactly  in 
the  former  measure.     This  exact  measure  is  the  greatest  com- 
mon measure  of  the  two  given  magnitudes. 

The  relation  of  a  to  an  may  be  expressed  thus: 
a  =maI+aa 
al=mIa2-\-a3 
a3=m3a3+a4 


.'.  The  ratio  of  a  toax  or  — —m-\ — -=m-\ — =w+- 
al  (/,  ax 


113.  By  a  similar  process  it  can  be  shown  that  two  magni- 
tudes may  not  have  a  common  unit  of  measure,  as  in  the  case 
of  the  side  and  the  diagonal  of  a  square.  We  will  show  that 

* means  and  so  on  according  to  the  same  lau\     It  is 

called  a  symbol  of  continuation. 


68 


Second-  Year  Mathematics 


if  A  B  (Fig.  90)  is  laid  off  on  A  C,  the  remainder  BjC  can  be 

laid   off    twice   on   B  C   leaving   a 


C 
B, 

A 


J)r £•         remainder    B2C.       This    can    be 

laid  off  twice  on   BjC,   leaving  a 
remainder  B3C. 

No   matter"  how  long  we  con- 
tinue this  process,  there  will  always 
be    a  remainder,   which,   however, 
can    be    made    indefinitely   small. 
pIG  Q0  This    is    shown   in   the    following 

outline  of  proof. 

Lay  off  (Fig.  91)  A  B  (=B  C)  on  A  C  to  Bx,  with  remainder 
BXC.  Draw  BIAI_LAC.  Then  BjC,  the  remainder,  equals 
BxA^A.B.  (Why?) 

Use  BjC  as  a  measure  of  B  C.  It 
is  contained  in  it  twice  to  B2  with  a 
remainder  B2C.  Draw  B2A2J_BC. 
Then  B2A2  =  A2BT  =B2C.  (Why  ?) 

Then  B2C,  the  last  remainder,  is 
contained  in  B^,  the  former  measure, 
twice  to  B3,  with  remainder  B3C. 

It  is  clear  that  this  process  can  be  FIG.  91 

continued    indefinitely,  each    measure 

after  the  first,  going  into  the  preceding  measure  twice  with  a 
remainder. 

A  C 

Thus  the  ratio  — -  cannot  be  found  exactly,  because  in 

D  C 

every  measurement  the  hypotenuse  of  an  isosceles  right 
triangle  is  measured  by  a  side,  and  there  is  always  a 
remainder. 

The  ratio  can,  however,  be  found  approximately  to  any 
degree  of  accuracy. 


Ratio,  Proportion,  Similar  Triangles  69 

Thus:     Since    A  C=B  C+BXC,    then    ^  =  1+^    or 

Jo  U  Jt>  ^ 

=-^  =  i  +  approximately. 
Jo  C 

B  C          B  C         AC 
Again,    since   B  C  =  2BIC+B2C  .'.  ^-^  =  2+B-^  ,  or  ^ 

i  AC 

=  i  + — v~F ,  and  15-^  =  1  +i  or  i  .5  approximately. 

±>2^  ±>    U 

2+BTC 
Again,    since    BIC  =  2B2C+B3C,    then   ^^  =  2+^-.'. 

D2(^  J52V^ 

AC  i  i 

=-^  =  iH or   iH — — r  or  1.4,  and  so  on  to  any 

B  C  i  2  +  5 


desired  degree  of  accuracy  not  absolute. 

1.4  is  an  approximate  value  of  1/2,  and  it  will  be  seen 
later  that  the  ratio  of  A  C  to  A  B,  or  B  C,  is  equal  to  j/2. 

114.  Two  magnitudes  which  have  a  common  unit  of  meas- 
ure are  said  to  be  commensurable.     Magnitudes  which  do  not 
have  a  common  measure  are  incommensurable. 

EXERCISES 

1.  Draw  three  segments  p  =  12  cm,  q= 6  cm,  r =4  cm.    Find 
a  common  unit  of  measure  and  check  by  applying  the  unit  to 
each  segment. 

2.  Given  three  segments  a,  b,  and  c,  such  that  a  can  be 
applied  to  b  m  times,  leaving  a  remainder  c;  c  can  be  applied 
to  a  n  times  leaving  a  remainder  d;    d  can  be  applied  to  c 
I  times  exactly.    What  is  the  ratio  of  a  to  6,  when  w=4,  n=2, 

Proportion 

115.  An  equation  whose  members  are  equal  ratios  is  a 


70  Second-Year  Mathematics 


i.  Are  the  following  statements  proportions: 
f  =  f?    f-  =  f?     J=lf?     Give  reasons. 


2.  Form  several  proportions. 

116.  Four  segments  are  in  proportion,  or  are  proportional, 
if  their  numerical  measures  are  in  proportion. 

117.  In  a  proportion,  as  k/l=m/n,  k,  I,  m,  and  n  are  the 
terms  of  the  proportion.     The  first  and  last  terms,  as  k  and 
n,  are  the  extremes.     The  second  and  third  terms,  as  I  and  m, 
are  the  means  of  the  proportion. 

1  1  8.  From  a  given  proportion  new  proportions  can  be 
formed  by  certain  rules.  Some  of  these  were  studied  in  First- 
Year  Mathematics,  pp.  125-30,  which  should  be  reviewed 
before  going  on  with  the  next  problems. 

Multiply  both  sides  of  the  proportion  a/b=c/d  by  bd. 
Taking  the  result  for  the  conclusion  and  the  given  proportion 
for  the  hypothesis,  the  following  theorem  is  obtained: 

PROPOSITION  I 

In  a  proportion,  the  product  of  the  extremes  is  equal  to  the 
product  of  the  means.^__ 

EXERCISES 

1.  Find  the  value  of  x  in  x/  2=12/3. 

2.  Let  ad=bc.     Prove  that  the  following  statements  are 
proportions  : 

(1)  a/b=c/d  (5)  b/a=d/c 

(2)  a/c=b/d  (6)  b/d=a/c 

(3)  c/a=d/b  (7)  c/d=a/b 

(4)  d/b=c/a  (8)  d/c=b/a 

To  prove  (i),  divide  both  members  of  equation  ad  =  bc  by  bd. 
119.  In  Problem  2,  the  following  theorem  is  proved: 


Ratio  ^Proportion,  Similar  Triangles  71 

PROPOSITION  II 

Theorem :  //  the  product  of  two  factors  is  equal  to  the  product 
of  two  others,  a  proportion  is  formed  by  taking  as  means  the  two 
factors  of  either  product,  and  as  extremes  the  two  factors  of  the 
other  product. 

EXERCISES 

1 .  Form  proportions  from  p3  —v3  =x2  —y2. 

2.  Form  proportions  from  pq=m2—n2. 

3.  Form  proportions  from  x2—y2=m3+n3. 

4.  Taking  (i)  of  Problem  2  (§118)  as  hypothesis,  prove  (2). 

5.  Taking  (3)  of  Problem  2  (§118)  as  hypothesis,  prove  (7). 

Notice  that  (2)  in  Problem  2  (§118)  may  be  obtained  from 
(i),  and  (7)  from  (3),  by  letting  the  means  interchange  places. 

This  process  of  interchanging  the  means  is  called  alternation. 

6.  Point  out  other  proportions  in  Problem  2  (§118),  which 
may  be  obtained  by  alternation  from  a  proportion  in  the  same 
problem. 

7.  Assuming  (i)  of  Problem  2  (§118),  prove  (5). 

8.  Assuming  (2)  of  Problem  2  (§118),  prove  (3). 

Notice  that  (3)  Problem  2  (§118),  may  be  obtained  from 
(2)  by  inverting  the  ratios  of  proportion  (2).  Therefore  (3) 
is  said  to  be  obtained  from  (2)  by  inversion. 

9.  What   proportion  is  obtained  by  inversion  from   (6)  ? 
from  (7)  ? 

PROPOSITION  III 

Theorem:  From  a  given  proportion,  proportions  may  be 
obtained:  (i),  by  interchanging  the  means  (alternation);  (2), 
by  interchanging  the  extremes  (alternation);  (3)  by  inverting 
both  ratios  (inversion). 


72  Second-Year  Mathematics    . 

EXERCISES 

1 20.  Solve  the  following  exercises: 

i.  In  Figs.  92  and  93,  the  segments  m,  n,  a,  6  are  in  pro- 
portion. Write  the  proportion  and  then  form  other  propor- 
tions by  alternation,  and  by 
inversion. 


a 


y 


FIG.  92  FIG.  93 

2.  Prove  the  theorem:   A  line  that  bisects  one  side  of  a  tri- 
angle, and  is  parallel  to  a  second  side,  bisects  the  third  side. 

Hypothesis:  A  D=D  B  (Fig.  94). 

DE||BC; 

Conclusion :    A  E = E  C . 
Proof:  Draw  E  F  j|  A  B. 
Prove  that  E  F=D  B  =A  D. 
Prove  triangle  A  D  E^F  E  C. 
ThenAE=EC. 

3.  Prove  A  D,  D  B,  A  E,  and  E  C  (Fig.  94)  are  proportional. 

4.  Since  E  F  ||  A  B 
(Fig.  94),  it  follows  that 
BF=FC.  (Why?)  (See 
Problem  2.) 

Prove  that  DE=JBC. 

jj  p  (j  5.  Prove  the  theorem: 

p  If  three  or  more  parallel 

lines  cut  two  transversals, 

and  if  the  segments  intercepted  on  one  of  the  transversals  are 
equal,  the  segments  intercepted  on  the  other  are  equal  also. 


Ratio,  Proportion,  Similar  Triangles  73 


x=y  (Fig.  95).     Why? 
Then  AI  =  AH. 
Therefore  A'B'  =  B'C'. 
Similarly  B'C'  =  C'D'. 


t 

7± 


t V 


7 


9-  \ 

FIG.  95 

6.  Prove  that  A  B,  A'B',  B  C,  B'C',  etc.,  are  proportional. 

7.  Prove  the  following  theorem:   A  line  drawn  through  the 
middle  point  of  one  of  the  sides  of  a  trapezoid  parallel  to  the 
bases,  bisects  the  other  side. 

PROPOSITION  IV 

Theorem:  //  two  parallel  lines  cut  two  intersecting  trans- 
versals, the  segments  of  one  transversal  are  proportional  to  the 
corresponding  segments  of  the  other. 

Hypothesis:  A  B  ||  AjB,  (Fig.  96). 
AT  A  and  BtB  in- 
tersect at  C; 


bVBUUUUU  . 

W  A  A, 

M  CA' 

CB 

A./ 

\» 

M      f 

\      p 

A/ 

vs 

(2)  CA, 

(  \ 

CBX 
BBX 

/ 

\ 

CA      CB  lG-9 

Proof  of  (i):  Take  A  Ax  as  a  measure  of  C  Ax.  It  will  be 
contained  an  integral  number  of  times,  say  twice,  with  prob- 
ably a  remainder,  A2C. 


74  Second-Year  Mathematics 

But  B  BT  will  measure  BZC  twice  and  leave  a  remainder, 
as  B2C  (Problem  5,  §  120). 

Use  the  remainder  A2C  as  a  measure  of  the  former  meas- 
ure A  A!. 

It  will  be  contained  in  it  an  integral  number  of  times  +  a 
remainder  as  A3At. 

But  B2C  will  go  into  B  Br  the  same  number  of  times  as  did 
A2C  in  A  A!  and  leave  a  remainder.  (Why  ?) 

If  this  process  is  continued  and  the  successive  divisions 
come  to  an  end  because  a  remainder  is  found  that  is  con- 
tained exactly  in  the  preceding  measure,  this  remainder  is  a 
common  measure  of  A  C  and  A  Ax  (§  112). 

But  we  have  also  seen  that  the  units  of  measure  of  AXC 
and  A  Ax  cut  B  C  by  lines  ||  to  A^  into  units  that  measured 
B^  and  B  Bz  the  same  number  of  times  as  did  the  units  of 
measure  of  AXC  and  A  Az. 

Consequently  the  numerical  measures  of  AXC  and  A  Ax  are 
the  same  as  those  of  B^  and  B  B!,  and  consequently  their 

ratios  are  equal.     That  is,  -T-^T-  =^-^-  . 

A  Aj      x>  tjI 

Prove  conclusions  (2)  and  (3)  in  like  manner. 

121.  If  the  lines  A  Az  and  AXC  are  incommensurable,  the 
successive  divisions  never  come  to  an  end  because  there  is 
always  a  remainder.  But  we  have  seen  (§  113)  that  the  ratios 
AXC  :  AXA  and  BTC  :  BjB  can  be  found  approximately  to  any 
desired  degree  of  accuracy,  and  at  every  approximation  the 
ratios  will  be  found  equal. 

The  two  ratios  thus  formed  are  then  equal,  since  if  they 
were  said  to  be  unequal,  differing  by  d,  we  could  prove  them 
to  differ  by  less  than  d  (see  also  "Approximation  of  Values  of 
Irrational  Numbers,"  First-Year  Mathematics,  p.  295). 


Ratio,  Proportion,  Similar  Triangles  75 

COROLLARY  I 

A  line  parallel  to  one  side  of  a  triangle  divides  the  other  two 
sides  proportionally. 

Prove  Proposition  IV  for  Fig.  97. 


Bl 


\ 

FIG.  97 

PROPOSITION  V 

Theorem:  //  any  number  of  parallel  lines  cut  two  trans- 
versals, the  segments  of  one  transversal  are  proportional  to  the 
corresponding  segments  of  the  other. 

ZtXV       ' 

&/  a/\Qfr 


\C 
- 


FIG.  98 
(See  Fig.  98.)     Prove  a=a',  b=b'.     Use  Proposition  IV. 

EXERCISES 

1.  Prove  Proposition  V  for  the  case  in  which  D  B  ||  E  C. 

2.  Prove:    Two  lines  that  cut  two  given  intersecting  lines, 
and  make  the  corresponding  segments  of  the  given  lines  pro- 
portional, are  parallel.     (Converse  of  Proposition  IV,  p.  73.) 


76  Second-Year  Mathematics 

It  is  assumed  that  the  cutting  lines  do  not  intersect  each  other 
between  the  given  lines. 

_       _     .        AB    ABt 
Hypothesis:     g-Q=g-Q~     (Flg- 99); 

Conclusion:     B  Bt  ||  C  C,. 


YT 

FIG.  99 

Proof   (indirect  method):    Suppose   E^x   not  parallel   to 
C  CT.     Then  C  C2  can  be  drawn  parallel  to  B  BT.     Therefore 

M=ABi       why?      But    AB=ABl      Why? 
BC    B.C/  BC    B^/ 

^liJ^-li.      Why?     /.B.C.-B^.     Why? 

rijL.a       ±5!^! 

This  is  impossible  (why?)  and  the  assumption  that  B  B, 
is  not  parallel  to  C  Ct  is  wrong.     Therefore  B  BT  ||  C  Cx. 

3.  Prove  (Fig.  99)  that,  if  ^|=^  ,  then  B  B,   ||  C  C,. 

4.  Prove  Problem  2  for  Fig.  100. 

X / 


/ 

FIG.  100 


Ratio,  Proportion,  Similar  Triangles 


77 


5.  Prove:  The  line  joining  the  middle  points  of  two  sides 
of  a  triangle  is  parallel  to  the  third  side  (the  converse  of  Prob- 
lem 2,  §  120). 


7 


7 


\ 


FIG.  ioi 

6.  Prove  that  the  line  through  the  midpoints  of  the  non- 
parallel  sides  of  a  trapezoid  is  parallel  to  the  bases  (Fig.  ioi). 

PROPOSITION  VI 

Theorem:  The  medians  of  a  triangle  meet  in  a  common 
point. 

Let  two  of  the  medians  meet  at  F  (Fig.  102)  and  prove  that 
the  segment  A  G  drawn  from  A  through  F  is  a  median. 

Draw  B  H  ||  D  F.    Then  A  F=F  H.    Why? 

Draw  H  C.    Then  E  F  ||  H  C.    Why  ? 

Figure  B  H  C  F  is  a  parallelogram.     Why  ? 

/.  B  G  =  G  C.  Why?  Hence,  A  G  is  the  median  of  tri- 
angle A  B  C  to  the  side  B  C. 

EXERCISES 

i.  Prove  in  Fig.  102  that  F 
is  a  trisection  point  of  the 
medians,  i.e.,  FG= JAG,  FE  = 
JBE,  FD=|DC. 

The  intersection  point  of  the 
medians  of  a  triangle  is  its  center 
of  gravity,  for,  if  supported  under 
that  point,  the  triangle  will  be 
found  to  balance. 


Second-  Year  Mathematics 


PROPOSITION  VII 

Theorem:  The  bisector  of  an  interior  angle  of  a  triangle 
divides  the  opposite  side  into  segments  whose  ratio  is  equal  to 
the  ratio  of  the  adjoining  sides  of  the  triangle, 

Hypothesis:    Z#  =  Zy     (Fig.  103); 


D 
FIG.  103 


Conclusion : 


AD    AB 


DC    BC' 
Proof:    Draw  A  E  ||  B  D. 
Prolong  C  B  to  meet  A  E  in  E. 
AD    EB 


Then 


DC     BC' 


(Why?)    Prove  Zz=Zw. 


Then  E  B  =A  B  (why  ?)  and 


AD    AB 


DC    BC' 


(Why  ?) 


EXERCISES 

1.  In  a  triangle  A  B  C,  A  B  =  5  in.,  B  C=4  in.,  A  C=3  in., 
A'B'  is  drawn  parallel  to  A  B  making  A'C=£  in.     Find  the 
lengths  of  all  segments  in  the  figure. 

2.  Prove  that  the  quadrilateral  whose  vertices  are  the  mid- 
points of  the  sides  of  a  triangle  and  one  vertex  of  the  triangle, 
is  a  parallelogram. 

3.  Prove  that  the  midpoints  of  the  sides  of  a  quadrilateral 


Ratio,  Proportion,  Similar  Triangles  79 

may  be  taken  as  vertices  of  a  parallelogram.     Prove  that  this 
parallelogram  is  equal  to  one-half  the  quadrilateral. 
4.  State  and  prove  the  converse  of  Proposition  VII. 

122.  A  point  P  on  a  segment  A  B,  divides  A  B  into  the 
parts  A  P  and  P  B. 

A B P 

FIG.  104 

The  segment  A  B  may  be  thought  of  as  described  by  a 
moving  point,  for  example,  by  letting  A  move  to  B  or  by  letting 
B  move  to  A.  The  segment  thus  has  two  directions.  If  one 
of  the  two  directions  of  a  segment  is  considered  positive,  the 
other  direction  is  negative.  Thus,  if  A  B  is  positive  (+),  B  A 
is  negative  (— ). 

Taking  a  point  P  on  a  segment  A  B,  then  (+AP)  + 
(+PB)  =  (+AB),  expresses  the  fact  that  the  whole  is  equal 
to  the  sum  of  its  parts.  If  P  is  on  the  extension  of  A  B,  as  in 
Fig.  104  (+AP)  +  (PB)  =  +  (AB),  since  P  B  is  negative, 
being  in  direction  opposite  to  that  of  A  P.  Thus  the  equa- 
tion holds  in  both  cases.  Because  the  sum  of  A  P  and 
P  B  is  A  B,  A  P  and  P  B  are  called  parts  of  A  B,  and  A  B 
is  said  to  be  divided  into  two  parts  externally  by  P.  Here 
the  two  parts  are  measured  one  from  A  to  P,  and  the  other 
from  P  to  B,  the  same  as  when  P  is  between  A  and  B. 

123.  Theorem:    The  bisector  of  an  exterior  angle  of  a  tri- 
angle divides  the  opposite  side  externally  into  segments  whose 
ratio  is  equal  to  the  ratio 

of  the  adjoining  sides  of 
the  triangle. 

Using  Fig.   105,  follow 
the  proof  of  Proposition  VII. 

State  and  prove  the  A 
converse  of  this  theorem. 


8o  Second-Year  Mathematics 

PROPOSITION  VIII 

If  two  or  more  ratios  are  equal,  the  sum  of  the  antecedents 
is  to  the  sum  of  the  consequents  as  any  antecedent  is  to  its  conse- 
quent. 

Hypothesis:  a/b=c/d=e/f=g/h=.  .  .  .  ; 


Conclusion:  "    .....  =a/b=c/d  =  .  . 

b+d+f+h+  ..... 

Proof:    a/b=a/b 

c/d=a/b(?) 
e/f=a/b(?) 
g/h=a/b(?)  ..... 
Therefore,  ab=ba(?) 
cb=da(?) 
<*-/*(?) 

gb=ha(?)  ..... 

Adding  (a-\-c+e+g+  .  .)b  =  (b+d+f+h  +  .  .)a.     (?) 
a+c+e+g+  .  .  .     a 


EXERCISES 

1.  If  a/b=c/d,  prove  that  r-r^=^/^- 

0  ~j~(t 

2.  Prove  that,  if  a/b=c/d,  ——z=a/c=b/d. 


3.  If  a/b=c/d,  prove  that 

a±b    c±d  a±b    c±d 


or  in  words:  In  a  proportion  the  sum  (or  difference)  of  the 
terms  of  one  ratio  is  to  the  antecedent  or  consequent  as  the  sum 
(or  difference}  of  the  terms  of  the  other  ratio  is  to  its  antecedent 
or  consequent. 


Ratio,  Proportion,  Similar  Triangles  81 

The  resulting  proportion  is  said  to  be  obtained  from  the 
given  proportion  by  addition  if  the  sum  is  taken,  and  by 
subtraction  if  the  difference  is  taken. 

The  names  composition  and  division  are  sometimes  given 
to  these  processes. 

T,     .,        , ,  .,    .  a  +  b    c+d 

4.  If  a  b=c  d,  prove  that r= -. . 

a—b    c—a 

x—a+b    a—b—x 


5.  Apply  addition  and  subtraction  to 


x+a  — b    a+b+x ' 


i—x 

6.  Apply  addition  and  subtraction  to  -          =3. 

V  i+x—  V  i—  x 


7.  Solve  for  x:  -  —  ^==—  —  -  -  —  .     (Ans.  x=—  i.) 
2—  V  x    VX+S  —  YX 


FIG.  1  06 


8.  In  triangle  ABC,          -          (Fig-  106). 

,       BA    BC          ,  BA    BC 
Prove  that          =—  ;   and 


. 

Problems  of  Construction 

124.  In  a  proportion,  as  a/b=c/d,  d  is  the  fourth  propor- 
tional to  a,  b,  and  c;  and  a  is  the  fourth  proportional  to  d,  c, 
and  b. 

i  .  To  construct  the  fourth  proportional  to  three  given  seg- 
ments. 

Given:    the  segments  A  B,  CD,  and  E  F. 


82 


Second- Year  Mathematics 


Required:  to  construct  the  fourth  proportional  to  A  B, 
C  D,  and  E  F. 

a)  Algebraic  solution:  Let  x  be  the  fourth  proportional. 
Measure  A  B,  C  D,  E  F,  and  substitute  these  numbers  in  the 

AB    EF 
proportion     ——. 


Solve  this  equation  for  x. 

Construct  a  segment  whose  measure  is  x.     This  is  the  re- 
quired fourth  proportional. 

b)  Geometric  solution  :    On  one  of  two  intersecting  lines,  as 
P  Q,  lay  off  P  S=A  B,  S  T  =  C  D  (Fig.  107). 

o  On    the    other,    as 

PR,  lay  off  PU=EF. 
Draw  SU. 

Draw  T  X  ||  S  U. 
Then    U  X    is    the 
required  fourth  propor- 


U 

FIG.  107 


X 


tional. 

Prove  by  Proposition  IV. 

2.  To  construct  the  product  of  two  numbers  m  and  n. 
Notice  that  i  :  m  =  n  :  mn. 

3.  To  construct  the  ratio  of  two  numbers  m  and  n. 
Observe  that  n  :  i  =  m  :  m/n. 

4.  Find  the  fourth  proportional  to  i,  2,  and  8;   to  pq,  pr, 
and  p. 

5.  Solve  for  *:    ^57=4/13. 

/c  c  i    c  P~Q  P+Q   P~Q 

6.  Solve  for  x:    x  :  - — -  =- — *  :  £ — *  . 

pr       p-q     p-q 

125.  In  a  proportion,  as  a/b=b/c,  c  is  called  the  third 
proportional  to  a  and  b,  and  a  is  the  third  proportional  to  c 
and  b. 


Ratio,  Proportion,  Similar  Triangles 


1.  Construct  the  third  proportional  to  two  segments 

(a)  Algebraically  (see  §  124). 
(6)  Geometrically. 

2 .  To  divide  a  segment  A  B  internally  and  externally  in 
the  ratio  of  m/n,  i.  e.,  (i)  to  find  a  point,  P,  on  A  B  so  that 
A  P/P  B=w/«,  and  (2)  to  find  a  point,  P',  on  the  extension 

A  P' 

of  A  B  so  that  ^T^-^W/W  (see  §  122). 

I.  To  divide  A  B  internally  in  the  ratio  m/n. 
Let  A  B  (Fig.  108)  be  the 

given  segment.  Draw  a  line 
A  C  through  A  and  lay  off 
A  D  =  m  and  D  E  =n.  Draw 
E  B.  Through  D  draw 
D  P  ||  E  B.  Then  P  divides 
A  B  internally  in  the  ratio 
of  m/n. 

II.  To  divide  A  B  externally  in  the  ratio  m/n. 

Draw  as  before  A  D  =m  (Fig.  109)  but  D  E=w.     Join  E  to 

AP'    m 


B  and  draw  D  P'  II  E  B.     Then 


P'B 


Prove. 


FIG.  109 

3.  A  segment  AB  =  i8  is  divided  internally  or  externally 

A  P 

at  a  point  P.     What  is  the  ratio  =r=  for  A  P  =  2  ?  3?  6?   9? 

r  D 

12?       2O?       24? 


84  Second-Year  Mathematics 

4.  Given  a  segment  AB  =  i8.     To  find  points  P  on  the 

AP    2     AP     .   AP  AP        3 

line  AB  so  that  —  --;    ^=6,  ^—4,          —  • 


5.  The  sum  of  two  unknown  segments  is  s  and  the  ratio 
is  the  same  as  the  ratio  of  two  given  segments  m  and  n.  Con- 
struct the  unknown  segments. 

126.  If  a  segment  is  divided  internally  and  externally  in 
the  same  ratio,  it  is  said  to  be  divided  harmonically. 

The  endpoints  of  the  segment  and  the  two  points  of 
division  are  called  four  harmonic  points. 

If  four  harmonic  points,  as  A,  B,  C,  D  (Fig.  no)  are  joined 
to  a  point  P  outside  of  line  A  B,  the  lines  P  A,  P  B,  P  C,  and 
and  P  D  form  an  harmonic  pencil. 


FIG.  no 

EXERCISES 

1 .  Prove  that  the  bisector  of  an  angle  of  a  triangle  and  the 
bisector  of  the  external  angle  having  the  same  vertex  divide 
the  opposite  side  harmonically. 

2.  Prove  that  if  the  segment  A  B  is  divided  harmonically 
by  points  C  and  D,  the  segment  C  D  is  divided  harmonically 
by  the  points  A  and  B. 

3.  To  divide  a  given  segment,  A  B,  into  segments  propor- 
tional to  several  given  segments,  x,  y,  and  z. 

On  a  ray,*  as  A  C  (Fig.  in)  lay  off  x,  y,  z  successively  and 
*  A  line  extending  in  one  direction  from  a  point,  is  a  ray. 


Ratio,  Proportion,  Similar  Triangles  85 

join  B  to  the  last  point  of  division  D.     Draw  parallels  to 
B  D  at  the  points  of  division.     Use  Proposition  V,  p.  75. 


B 


FIG.  in 


4.  To  divide  a  segment  into  equal  parts  (Fig.  112). 

Construction  is  the  same  as 
in  Problem  3,  using  equal  seg- 
ments instead  of  x,  y,  and  z 
in  Problem  3. 

5.  A  segment  A  B  =30  cm. 
A  point  P  is  10  cm  from  A. 
Determine  a  point  P'  so  that  A, 
FIG.  112  P,  B,  P7,  are  harmonic  points. 

6.  The  segment  XY  is  divided  harmonically  by  the  points 

i         /i         i  \ 
A  and  B.    Prove  that  ™=i  ^FT  +  ^5     (FiS-  IJ3)- 


Proof: 

XA 

XB 

X 

A 

y 

B 

AY 

BY 

FIG.  113 

AY 

BY 

(Why?) 

XY-XA    XB 

-XY 

the 

"XA 

XB' 

XA 

XB        ^ 

figure). 

XY 

XY 

XY 

XY 

-1=1 

•''2=XA  + 

"XA 

XB' 

XB 

=  +         '  dividinS  both  sides' 


86  Second-Year  Mathematics 


7.  The  segment  XY  is  divided  harmonically  by  the  points 
A  and  B.     Prove  that 


i       ,  /   i         i   \ 


BY/  ' 

Method  of  Analysis 

127.  When  no  obvious  solution  of  a  problem  presents 
itself,  the  following  method  may  be  used  to  discover  the  solu- 
tion: 

(1)  Assume  that  the  problem  is  true,  and  consider  results 
which  might  follow.     Continue  deducing  other  results  from 
these  results,  until  a  known  proposition  or  statement  is  reached 
(analysis) . 

(2)  Then  reverse  the  process,  thus  obtaining  the  solution 
(proof). 

This  method,  called  the  method  of  analysis,  is  often  useful 
in  attacking  problems  and  original  propositions. 

1.  Prove  that  if  a  :  b=c  :  d,  then  (a+b)  :  b  =  (c+d)  :  d. 

Analysis: 

a+b     c+d 

(1)  Assume  ~~r~=~j~  • 

0  CL 

(2)  Then  ad+bd  =  bc  +  bd. 

(3)  ad=cb. 

(4)  a:b=c:d. 

Proof: 

(1)  a:b=c:d. 

(2)  ad=cb. 

(3)  ad+bd=cb+bd.    (?) 

(4)  d(a+V)=b(c+d),    (?) 

(5)  (a+b):b  =  (c+d):d.  (?)     Q.  E.  D. 

2.  If  a/b=c/d,  prove  the  four  following  relations: 

(i)  c2/d2=ca/bd. 


(2) 


Ratio,  Proportion,  Similar  Triangles  87 

a  +  c    a2d 


b+d    b*c 


aa+ca==ab+cd 
w>  a»-c*~ab-cd' 

.  a+b+c+d    b+d 

W.^THT  :-^~ 

3.  If  a/b=b/c  prove  that 


a2+62     62+c2 
(2) 


(3) 


a  c 

a+c  _a  —  c 

ba+ca~~ba-c*' 

aa+ab    ba+bc 


a  c 

4.  If  a/b=c/d=e/f,  prove  that 

.  a2     ba 


(2) 

la—me+nc_a 
(3)  lb-tnf+nd  =  b' 

Algebraic  Exercises 
Solve  the  following  exercises  algebraically: 

1.  The  segments  of  the  lines  of  a  pencil,  or  ray  of  lines, 
from  P  made  by  a  pair  of  parallels  may  be  designated   as 
shown  in  Fig.  114.     Find  x,  y,  and  the  lengths  of  the  segments 
included  by  the  parallels. 

2.  A  pencil  like  that  of  Fig.  114  is  cut  by  two  parallels, 
so  that  PA=4,  PB  =  3,  PC=s,  PD  =  7,  and  AA'  =  i-;y, 
B  W=x  —  i,  C  C'=2X— 3,  and  D  D/  =  2^+i3. 

Find  x,  y,  and  the  last  four  segments  named. 


88 


Second-  Year  Mathematics 


3.  As  in  Fig.  114,  if  PA=3,  PB=PC  =  i,  PD  =  n,  and 
y-x,   EW=x+y-2,    CC'=3-y~3X,    and   D  D'  = 
2y— x,  find  x,  y,  and  the  last  four  segments. 


FIG.  114 

4.  Two  intersecting  lines  are  cut  by  a  pair  of  parallels  so 
that  segments  may  be  designated  as  shown  in  Fig.  115.  Find 
x  and  the  designated  segments. 

Can  you  give  any  geometrical  meaning  to  the  negative 
value  of  x  ? 


FIG.  115 

5.  If  the  parallels  cut  so  that  PA =9—  4*,  A  A7 =3—4^, 
PB  =  2o#— 3,  and  BB/=3—  4X,  find  x  and  the  segments  of 
the  non-parallels.     Draw  a  figure  to  illustrate  for  both  values 
oi  x. 

6.  With  the  segments  in  the  same  order  as  in  Exercise  5 
denoted  by  3^—1,  J—x,  8  +  2X,  and  2X  +  i,  find  x  and  the 
four  segments. 


Ratio,  Proportion,  Similar  Triangles  89 

Similar  Triangles 

128.  Similar  triangles  have  been  denned  (see  §§  82-84, 
First-Year  Mathematics)  as  follows: 


Two  triangles  having  the  corresponding  angles  equal  and 
the  ratios  of  the  corresponding  sides  equal,  are  similar. 

129.  Similar  polygons  are  denned  in  the  same  way.     Then 
to  prove  that  two  polygons  are  similar,  we  must  show  that  the 
corresponding  angles  are  equal  and  that  the  corresponding  sides 
are  in  proportion.     Corresponding  angles,  sides,  or  other  lines 
are  often  called  homologous  angles,  sides,  or  lines. 

To  prove  that  two  triangles  are  similar,  however,  it  is 
sufficient  to  prove  that  only  some  of  these  conditions  are 
satisfied. 

It  will  be  shown  that  two  triangles  are  similar  under  any 
one  of  the  following  three  conditions: 

(1)  If  two  angles  of  one  triangle  are  equal  each  to  each 
to  two  angles  of  the  other. 

(2)  If  the  ratio  of  two  sides  of  one  triangle  equals  the  ratio 
of  two  sides  of  the  other  and  the  angles  included  by  these 
sides  are  equal. 

(3)  If  the  corresponding  sides  are  in  proportion. 

130.  Many  problems  can  be  solved  by  the  aid  of  similar 
triangles. 

PROBLEMS 

i.  Surveyors  use  similar  triangles  to  lay  off  parallel  lines. 
Let  A  C  be  a  given  line.  /^^ —rC 

To  construct  a  line  parallel    Zh ~I^u 

to  AC. 


Lay  off  a  triangle,  as  A  B  C 
(Fig.  1 1 6).  FIG.  116 


9o 


Second-  Year  Mathematics 


On  A  B  mark  a  point,  as  D.     Determine  a  point  E  on 

A  B     C  B 

C  B,  so  that  =-rjr  =:prw  •     Draw  D  E.     This  is  the  required 
D  B     Jb  r> 

parallel. 

Triangles  D  B  E  and  A  B  C  are  then  similar  to  each  other 
by  case  (2),  §  129. 

Therefore  Z.x=  Z.y  by  definition  of  similar  triangles  and 
D  E  ||  A  C. 

2.  A  perpendicular  to  a  line  from  a  remote  point  may  be 
located  by  similar  triangles. 

Let  A  be  the  point  and  B  C 
the  line  to  which  the  perpen- 
dicular from  A  is  to  be  drawn 
(Fig.  117).  Draw  AB.  From 
a  convenient  point  D  on  A  B 
draw  D  EJ_B  C. 

If  A  F  represents  the  required 
perpendicular  it  is  seen  that  triangle  B  E  D^  triangle  B  F  A 
case  (i),  §  129.  Therefore  F  can  be  located  by  the  fourth 

DB     EB 

proportional  in  ^~g=p-g  . 

PROPOSITION  IX 

Theorem:    Two  triangles  are  similar  if  two  angles  of  one 
are  equal  to  two  angles 
of  the  other. 

Hypothesis:     x=x'. 
y=/(Fig.  118); 

Conclusion:  A  ABC 
w~  AA'B'C'. 

Proof:  Prove  C=C'. 

Lay  off  CD=C'A', 
andCE  =  C'B'.  FIG.  118 


Ratio,  Proportion,  Similar  Triangles 


Draw  D  E.     Prove  AD  E  C 
Prove  x"=x. 


Then  D  E  ||  AB,,  . 


=        and 


C/A'    C/B/ 


Similarly  prove 


CA    CB 
C'B'    B'A 


j 
-          and          = 


CA"CB 
C'A' 


(Why?) 


i  .  Problem  :    To  find  the  height  of  a  chimney. 

Let  A'B'  be  a  vertical  stick  and  A'C'  its  shadow  (Fig.  119). 


C    A' 


FIG.  119 
Let  A  C  be  the  shadow  of  the  chimney. 

Prove  that  A  B  =  A'B'  •  ^  . 
A  (^ 

What  is  the  height  of  the  chimney  if  A'B' =4  ft.,  A'C'  = 
9ft.,  AC  =  io8ft.? 

EXERCISES 

1.  Triangles  having  the  corresponding  sides  parallel,  or 
perpendicular,  are  similar.     Prove. 

2.  If  lines  are  drawn  joining  the  midpoints  of  the  sides 
of  a  triangle,  another  triangle  is  formed  which  is  similar  to 
the  first  triangle. 

AB 
.AB 


3.  In  triangle  ABC  (Fig.  120)  j  ^  Jjj 


.V      .  CF 

ProvethatED  =        ' 


Second-  Year  Mathematics 


4.  The  bases  of  a  trapezoid  are  18  and  24,  and  the  altitude 
is  12.     Find  the  area. 

Produce  the  non-parallel  sides 
until  they  meet  and  compare  the 
altitudes  of  the  similar  triangles. 

5.  The  shadow  cast  by  a  4-ft. 
vertical  rod  is  5^  ft.,  and  the 
shadow  cast  by  a  spire  is  220  ft. 
How  high  is  the  spire  ? 


FIG.  120 


6.  A  man  at  a  window  sees  a  point  on  the  ground  in  line 
with  the  top  of  a  post  2  ft.  8  in.  from  the  foot  of  the  post  and 
finds  that  the  post  is  3  ft.  high,  and  24^  ft.  from  a  point  just 
under  the  window.     How  high  is  the  window  from  the  ground  ? 

7.  A  boy  wishes  to  know  how  far  it  is  from  the  shore  of 
a  lake  to  an  island,  and  makes  the  measurements  shown  in 
Fig.  121,  also  D  B  =  io  rd.     Find  the  required  distance. 

8.  The  boy  now  wishes 
to  lay  off  a  half-mile  course 
having  the  island  A  as  one 
end.     How  long  must  D  E 
be   made   that  the  line  of 
vision  from  B  to  E  shall 
cut  a  line  parallel  to  D  E 
through  A  at  a  point  J  mile 
from  A  ? 

9.  Make     a     drawing 
which  shall  represent  two 

stakes  in  a  lake  or  river,  and  lines  measured  on  the  land  by 
which  to  find  the  distance  between  the  stakes. 

10.  Make  a  similar  drawing  and  show  how  to  locate  the 
stakes  a  given  distance  apart. 

11.  The  non-parallel  sides  of  a  trapezoid  of  bases  60  and 
1  8  and  of  altitude  6,  are  produced  until  they  meet.     What 


B 


FIG.  121 


Ratio,  Proportion,  Similar  Triangles 


93 


are  the  altitudes  of  the  triangles  formed  on  the  bases  of  the 
trapezoid  ? 

12.  The  base  of  a  triangle  is  72  in.,  and  its  altitude  is 
12  in.     Find  the  upper  base  of  the  trapezoid  cut  off  by  a  line 
parallel  to  the  base  of  the  triangle  and  8  in.  from  it. 

13.  The  bases  of  a  trapezoid  are  nj  and  5,  and  the  non- 
parallel  sides  are  5^  and  3.     Find  the  sides  of  the  smaller  tri- 
angle formed  by  producing  the  sides  of  the  trapezoid. 

PROPOSITION  X 

Theorem:  Two  triangles  are  similar  if  the  ratio  of  two 
sides  of  one  equals  the  ratio  of  two  sides  of  the  other,  and  the 
angles  included  between  these  sides  are  equal. 

AC      C  B 
Hypothesis:    x=x'  and          =    '' 


FIG.  122 

Conclusion:    A  A  B  C^  AA'B'C'. 

Proof:    Lay  off  C'D-C  A,  C'E  =  C  B.     Draw  D  E. 

I  C'.     (Why  ?) 


DC'    EC' 
AfC'^Cm'     (Why?) 
.'.  D  E  ||  A'B'. 

Prove  AD  E  C^  AA'B'C'. 
C^  AA'B'C'. 


94  Second-Year  Mathematics 

EXERCISES 

i.  To  determine  the  distance  across  a  river. 

Sighting  across  the  river  with  telescope  A  (Fig.  123),  place 
rods  at  B  and  C  in  the  line  of  sight.  Take  readings  of  rods  at 
E  and  D.  Depress  the  telescope  sighting  at  C  and  take  a 
reading  at  F.  From  the  rod  readings  find  D  F  and  E  C. 


FIG.  123 

Since    EC  ||  DF,    /^DAF-AEAC,   and 
F  P 

,.AE-AD.|f. 

E  D  can  be  found  by  subtracting  A  D  from  A  E. 

2.  Two  isosceles  triangles  are  similar,  if  an  angle  in  one 
is  equal  to  the  corresponding  angle  in  the  other.     Prove. 

3.  Two  right  triangles  are  similar  if  the  ratio  of  the  sides 
including  the  right  angle  in  one,  is  equal  to  the  ratio  of  the 
corresponding  sides  in  the  other.     Prove. 

4.  Two  sides  of  a  triangle  are  14  and  3.5  in.  and  the  in- 
cluded angle  is  75°.     Two  sides  of  another  triangle  are  20  and 
5  and  the  included  angle  is  75°.     Are  the  triangles  similar? 
Give  reason  for  your  answer. 

5.  Prove  that  two  parallelograms  are  similar  if  they  have 
two  corresponding  angles  equal,  and  the  including  sides  in 
proportion. 

6.  Two  rectangles  are  similar  if  the  ratio  of  two  adjacent 
sides  of  one  is  equal  to  the  ratio  of  the  corresponding   sides 
of  the  other. 


Ratio,  Proportion,  Similar  Triangles  95 

PROPOSITION  XI 

Theorem:     Two  triangles  are  similar  if  the  corresponding 
sides  are  in  proportion. 

AB     BC     CA 

Hypothesis:    _-_  =  _  (Fig.  124); 


FIG.  124 

Conclusion:    A  A  B  C^  AA'B'C7. 

Proof:  Layoff  CD  =  C'A' 
andCE  =  C'B'. 

DrawDE.     Prove  AD  C  E^AA  C  B,  §  129.  (i) 

A  B    B  C 

:  =7=TF  •  (Similar  triangles  have  their  sides  in  proportion.) 


AB     BC 

A7!"'  =B7c>  *   (  y  hyp°thesls-) 

,       f       A  B  _  A  B       (Magnitudes  equal  to  equal  things 
A'B'    D  E  '         are  equal  to  each  other.) 


Prove  AD  C  E^  AA'B'C'.  (2) 

Therefore  AA  B  C^  AA'B'C'  (from  (i)  and  (2)). 

PROPOSITION  XII 

Theorem:     The  perimeters  of  similar  triangles  are  to  each 
other  as  any  two  homologous  sides. 

Denoting  the  sides  of  two  similar  triangles  by  a,  b,  c  and 


Second-Year  Mathematics 


a',  b',  c',  respectively;    we  have  a/a'=b/b'=c/cf.    Then  by 

a+b+c       a      b      c 

Proposition  VIII,  p.  80,         ,          =  —  =  7-  =  -.  • 
a'+b'+c'     a'     b'     c' 

EXERCISES 

1.  The  perimeter  of  a  triangle  is  15  cm.,  and  the  sides  of 
a  similar  triangle   are  4. 5  cm.,  6. 4 cm.,  and  7.1  cm.     Find 
the  lengths  of  the  sides  of  the  first  triangle. 

2.  A  race  course  in  the  shape  of  an  equilateral  triangle  is 
\  mile  long.     Find  the  sides  of  a  similar  course  80  ft.  long; 
1,760  ft.  long;  i  mile  long. 

3.  The  perimeters  of  two  similar  triangles  are  #2+3#  +  a 
and  1 6,  and  a  pair  of  homologous  sides  are  respectively  yx 
and  8.     Find  the  value  of  x. 

4.  The  perimeters  P  and  P'  of  two  similar  triangles,  and  a 
pair  of  homologous  sides  a  and  a'  may  be  expressed  in  the  fol- 
lowing table.     Find  the  values  of  x,  y,  and  k. 


0=35 
a=x  —  2 

a=x+i 


a'  =  2 
a' =4 
a' =6 
' 


0'=3 


FIG.  125 


0  =  5 

a =40  a'  =  2X  —  b 

5.  The  homologous  alti- 
tudes of  two  similar  tri- 
angles are  to  each  other  as 
their  homologous  sides,  or 
as  their  perimeters. 

Show  that  the  triangles 
T  and  T'  (Fig.  125)  are 
similar. 


Ratio,  Proportion,  Similar  Triangles 


97 


6.  The  homologous  medians  of  two  similar  triangles  are  to 
each  other  as  any  two  homologous  sides,  and  as  the  perim- 
eters. See  Fig.  126.  (amx^a'm'x'.}  Prove. 


c' 


FIG.  126 


PROPOSITION  XIII 

The  perpendicular  to  the  hypotenuse  from  the  vertex  of 
a    right    triangle     divides  £ 

the  triangle  into  triangles 
similar  to  each  other,  and 
to  the  given  triangle. 


x=x'  (?)     (Fig.  127) 


z=z' 


y   % 


z' 


FIG.  127 


.'.  AA  D  C^  AB  D  C^  AA  B  C. 

131.  In  the  proportion  a/b=b/c,  b  is  called  a  mean  pro- 
portional between  a  and  c. 

EXERCISES 

1 .  What  is  a  mean  proportional  between  4  and  9  ? 
Denoting    a    mean    proportional    by   x,   we   have   ^/x=x/g.  (?) 

x1  =  4  •  9  (  ?),  x=  ±  6.     Prove  both  values  correct. 

2.  What  are  mean  proportionals  between  2  and  18?    10 
and  90?  8  and  200?   20  and  180? 

3.  Find  a  mean  proportional  between  a3  and  b2;  between 
c2  and  d2. 

4.  Find  a  mean  proportional  between  x2  +  2xy-\-y2   and 
x2  —  2xy+y2. 


98 


Second-  Year  Mathematics 


132.  Since  triangle  ADC  (Fig.  127),  is  similar  to  triangle 

DEC,  it  follows  that  ?—^=^-E.     That  is— 
JJ  U     D  r> 

In  a  right  triangle,  the  perpendicular  from  the  vertex  of  the 
right  angle  to  the  hypotenuse  is  a  mean  proportional  between 
the  segments  of  the  hypotenuse. 

This  affords  a  way  of  finding  geometrically  a  mean  pro- 
portional between  two  numbers. 

PROBLEMS 

i.  Construct  a  mean  proportional  between  any  two  segments. 
Let  a  and  b  be  the  given  segments.     Required  to  construct 
a  mean  proportional,  x  (Fig.  128). 


X*  ' 

*"  v. 

x 

/ 

\ 

I/ 

\ 
\ 

/         a 

b     \ 

A                     £            C 

L 

FIG.  128 

On  any  line,  as  A  L,  lay  off  A  E=a  and  B  C=b. 

At  B  draw  B  D±A  C. 

Draw  a  circle  on  A  C  as  a  diameter,  meeting  B  D  at  D. 
Then  B  D  is  the  required  mean  proportional. 

Proof:  Assume  any  angle  inscribed  in  a  semicircle  to  be  a  R.  A. 
(proved  in  §  163,  2)  .'.  Z  A  D  C  =  i  R.  A.,  then  see  §132. 

2.  Construct  a  mean  proportional  between  25  and  16. 

3.  The  mean  proportional  between  a  and  b  is  the  square 
root  of  the  product  of  a  and  b.     Prove. 

Problems  3  and  i  suggest  a  way  of  finding  the  square  root 
of  a  number  geometrically. 

4.  Construct  the  square  root  of  6  to  two  decimal  places. 

On  squared  paper  lay  off  in  the  same  way  as  a  and  b 


Ratio,  Proportion,  Similar  Triangles 


99 


(Problem  i)  two  factors  of  6,  as  3  and  2,  to  the  scale  of  i  =sum 
of  the  sides  of  10  small  squares.  Erect  a  perpendicular  as  x  in 
Problem  i,  and  measure  this  perpendicular. 

5.  Find  geometrically  to  two  decimal  places  the  square 
root  of  8. 

6.  Find  geometrically  to  two  decimal  places  the  square 
root  of  5. 

7.  In  a  right-angled  triangle  either  side  including  the  right 
angle  is  a  mean  proportional  between  the  hypotenuse  and  the 
adjacent  segment  of  the  hypotenuse,  made  by  the  perpendicular 
from  the  vertex  of  the  right  angle  to  the  hypotenuse. 

Proof  follows  from  similarity  of  the  triangles  ABC  and  A  C  D  and 
of  triangles  A  B  C  and  D  B  C  (Proposition  XIII). 

8.  In  triangle  ABC  ZC  =  i  R.  A.,  C  D±A  B.     A  D  =  2, 
D  B  =30.     Find  the  lengths  of  A  C  and  C  B  (Fig.  129). 

The  segment  of  the  hypotenuse  adjacent  to  either  side  is 
called  the  projection  of  that  side. 

9.  The  radius  of  circle  O  is  12.5  and  a  chord,  as  AC, 
equals  15.     Find  the  projection  of  A  C  upon  the  diameter 
through  A  (Fig.  130). 

C 


30 


AD  -B 

FIG.  129 

Similar  Polygons 
PROPOSITION  XIV 

Theorem:     Homologous  diagonals  divide  similar  polygons 
into  similar  triangles. 

Hypothesis:  AB  C  D  .  .  .  .  ^A'B'C'D'  ....  (Fig.  131); 
Conclusion:      \I^AI 


100 


Second-  Year  Mathematics 
CD      DE 


•'.?=/(?) 

CE     EF  ,  DE 

^Er,=^r^f  as  both  ratios  equal  . 

Therefore  AH  —  A  IF,  etc 

EXERCISES 

133.  Solve  the  following  exercises: 

i.  The  perimeters  of  similar  polygons  are  to  each  other  as 
any  two  homologous  sides. 

Denoting  the  sides  of  the  polygons  by  a,  b,  c,  .  .  .  .  and  a',  b',  c' 

a      b      c 

.  .  .  .     then— ,  =  —  =  —  = Why? 

a      b      c 

a+b+c ....        a      b      c 


a'  +  b'  +  c' a'     b'     c'' 

2.  State  and  prove  the  converse  of  Proposition  XIV. 
The   ZBCA   (Fig.   132)  being  a  right  angle,  show  the 

following : 

3.  c  :  a=a  :  m  or  that  a2=m  •  c. 


Ratio,  Proportion,  Similar  Triangles 


101 


4.  c  :  b=b  :  n  and  b*=n  •  c. 

5.  aa-\-ba  =  (m+n)c—ca    (Add  Ax.    applied    to    Exercises 
3  and  4). 

6.  And  a2/b2=m/n.     (Div.  Ax.). 

7.  Also  aa=ca-b3  and  b2=c*-a2.     (?) 

8.  It  has  been  shown  that  m:h=h:n  or  h  =  -\/mn.     (?) 

9.  From  Exercise  5  and  Exercise  7  it  follows  that  in  a 
right  triangle  the  square  of  the  hypotenuse  is  equal  to  the  sum 
of  the  squares  of  the  other  two  sides,  and  the  square  of  either 
side  about  the  right  angle  is  equal  to  the  square  of  the  hypote- 
nuse less  the  square  of  the  other  side. 

10.  a  =  i2and&  =  5.    Find  C 
c,  m,  n,  and  h     (Fig.  132). 

11.  a=8andc  =  io.    Find 
b,  m,  n,  and  h. 

12.  w=9|      and     «  =  5l- 
Find  a,  b,  c,  and  h. 

Solve  Exercise  12  first  for 
h  by  Exercise  8,  then  for  a  by  Exercise  5,  aa=ma+ha,  and 
check  results  in  Exercises  3,  4,  and  6. 

13.  Find  geometrically  a  line-segment  whose  length  squared 

shall  be  f  of  the 
square  of  the  length 
of  a  given  segment 
(i.e.,  the  ratio  of  their 
squares  shall  be  5 : 3). 
Draw  a  line  as  A  B, 
and  divide  it  into 


»A  segments  =  5     to     3 
(Fig.  133). 


8 

FIG.  133 

Draw  semicircle, 
and  perpendicular.     Lay  off  C  D  the  length  of  the  given  line. 


102 


Second-  Year  Mathematics 


Draw  D  E  ||  A  B.     C  E  is  the  required  line. 

Proof:    CE  :  CD  =  CB  :  C  A.  (?) 

CE2  :  CD2=CB2  :  C  A2.  (?) 

But  C  B2  :  C  A2=w  :  n  =  $  :  3  (see  Exercise  6). 

CE2:CD2  =  5:s.     Hence  C  E2=f  C  D2. 

14.  Solve  Exercise  13  when  C  D  is  longer  than  C  A. 

15.  Solve  Exercise  13  when  the  ratio  is  7  :  4. 

Similar  Right  Triangles,  Trigonometry 
134.  Solve  the  following  problems: 

1.  All  right  triangles  having  an  angle  of  30°  are  similar. 
Why? 

2.  On  squared  paper  draw  a  right  triangle  having  an  angle 
of  30°.     Measure  the  sides,  and  find  the  ratio  of  the  side 
opposite  the  angle  30°  to  the  hypotenuse. 

Prove  that  this  ratio  is  the  same  for  all  right  triangles 
having  an  angle  of  30°. 

3.  In  the  triangle  of  Problem  2,  find  the  ratio  of  the  side 
opposite  the  angle  60°  to  the  hypotenuse. 

Prove  that  this  ratio  is  constant  for  all  right  triangles 
having  an  angle  of  60°. 

4.  In  a  right  triangle  having  an  angle  of  45°,  find  the  ratio 
of  the  side  opposite  the  angle  45°  to  the  hypotenuse. 

Prove    that    this 

Mr  ratio  is  constant  for  all 

right  triangles  having 
an  angle  of  45°. 

5.  Draw  with  a 
protractor  an  angle 
of  40°.  From  any 
points,  as  Ax,  A2,  A3, 
A4  .  .  .  .  (Fig.  134) 


Ratio,  Proportion,  Similar  Triangles  103 

on  either  side  of  the  angle,  draw  perpendiculars  to  the  other 
side.     Measure  A^,  and  AtO  and  find  their  ratio. 

6.  Prove  that  the  ratio  of  the  side  opposite  the  angle  40° 
to  the  hypotenuse  is  the  same  for  all  triangles  of  Fig.  1 34. 

135.  The  constant  ratio  of  the  opposite  side  to  the  hypote- 
nuse as  in  Fig.  1 34,  is  called  the  sine  of  angle  40°. 

1.  Find  the  sine  of  70°;   of  50°;  of  20°. 

2.  On  square-ruled  paper,  draw  an   acute   angle  whose 
sine  is  £.     Find  the  number  of  degrees  in  the  angle. 

3.  Find  the  number  of  degrees  in  an  acute  angle  whose 
sine  is  |;    .2;    .75. 

136.  Many  problems  in  science,  and  in  surveying  and  other 
branches  of  engineering,  are  solved  by  the  aid  of  the  ratios  of 
the  sides  of  right  triangles. 

Let  AOD  (Fig.  135) 
represent  any  acute  angle. 
Construct  a  right  triangle 
containing  the  angle  AOD, 
by  drawing  from  any  point 

P,  on  either   side,   a  line       ^ 

PM  perpendicular  to  the    0  ft  P        D 

other  side.  FlG-  J35 

The  ratio  of  M  P  to  O  P  is  called  the  sine  of  angle  O 
(written  sin  O). 

The  ratio  of  M  O  to  O  P  is  called  the  cosine  of  angle  O 
(written  cos  O). 

The  ratio  of  M  P  to  O  M  is  called  the  tangent  of  angle  O 
(written  tan  O). 

EXERCISES 

i .  Prove  that  the  sine  of  a  given  acute  angle  O  is  the  same 
for  all  right  triangles  that  have  the  angle  O,  i.  e.,  that  the 
sine  of  O  does  not  depend  upon  the  size  of  the  right  triangle 
that  contains  the  angle  O. 


IO4 


Second-Year  Mathematics 


2.  Prove  that  the  cosine  of  a  given  acute  angle  is  constant. 

3.  Prove  that  the  tangent  of  a  given  acute  angle  is  constant. 

137.  It  is  customary  to  denote  the  angles  of  a  triangle  by 
capital  letters,  and  the  sides  opposite  by  the  corresponding 
small   letters.     Let   ABC    (Fig.  1 36)  be  any  right  triangle 
(C  being  the  right  angle),  and  let  a,  b,  c  denote  the  lengths  of 
the  sides  opposite  the  angles  A,  B,  C,  respectively. 

The  definitions  of  §136  may  then  be  stated  briefly  as  follows: 

a    opposite  side 
sin  A=-=-p- 

c      hypotenuse 

b    adjacent  side 
cos  A  =-  =  -T^-  —  . 

c      hypotenuse 

a    opposite  side 

tan  A=T=-^T — 7-r-  . 

b     adjacent  side 

138.  It  is  important  to  note  that  the  sine,  cosine,  and  tan- 
gent of  an  angle  are  numbers,  the  ratios  or  quotients  of  certain 
lengths. 

The  ratios  have  definite,  fixed  values  for  any  given  angle, 
i.  e.,  they  are  constant  numbers  for  a  given  angle. 

The  sine,  cosine,  and  tangent  of  an  angle,  as  defined  in 
§137,  for  acute  angles  cannot  be  negative,  since  no  one  of  the 
numbers  a,  b,  c  (Fig.  136)  is  negative. 


i.  Give  the  sine,  cosine 
(Fig.  136). 


PROBLEMS 
:,  and  tangent  of  the  acute  angle  B 


Ratio,  Proportion,  Similar  Triangles 


2.  In  Fig.  136,  show  that 

(1)  sin  A=cos  B 

(2)  cos  A=sin  B 


(3)  tanA  = 


i 


tan  B 


sm  A 

(4)  tan  A  = . 

cos  A 


3.  Show  that  the  sine  of  an  angle  is  less  than  (or  at  most 
equal  to)  i. 

In   Fig.    136,   c*=a*+b2    (§133,   Problem   9,  and    §142, 
First-Year  Mathematics). 
Hence,  a <c.     Why? 
Therefore,  sin  A  =a/c<  i .    Why  ? 

4.  Show  that  the  cosine  of  an  angle  is  less  than  i . 

5.  On  square-ruled  paper,  with  a  protractor,  draw  angles 
of  40°,  50°,  25°,  65°.     Find  by  measuring  the  lines,  the  sine, 
cosine,  and  tangent  of  each  angle  and  compare  the  results 
with  the  values  in  the  following  table : 


25° 

s°° 

40° 

65° 

Sine  

.4226 

.7660 

.6428 

.0063 

Cosine  

.0063 

.6428 

.  7660 

.4226 

Tangent   

.466? 

i  .  1918 

•8^01 

2  .  I441} 

6.  The  lengths  of  the  sides  of  a  right  triangle  are  repre- 
sented by  a,  b,  and  c,  the  hypotenuse  being  c.     Find  the  sine, 
cosine,  and  tangent  of  angle  A  and  of  angle  B  under  each  of 
the  following  conditions: 

(1)  a=3,  6=4  (3)  6  =  15,  c  =  i7 

(2)  a  =  s,  £  =  13  (4)  a  =  7,    £  =  25. 

7.  On  square-ruled  paper  draw  the  angle  whose  sine  is  £. 
Find  the  cosine  and  tangent  of  the  angle,  and  the  number  of 
degrees  in  the  angle. 

8.  Draw  the  angle  whose  cosine  is  £.     Find  the  sine  and 
tangent  of  the  angle,  and  the  number  of  degrees  in  the  angle. 


io6 


Second-  Year  Mathematics 


9.  Draw  the  angle  whose  tangent  is  i.     Find  the  sine  and 
cosine  of  the  angle,  and  the  number  of  degrees  in  the  angle. 

10.  Draw    an   equilateral    triangle 
(ABC,  Fig.  137). 

Prove  that  the  altitude  B  H  bisects 
A  C  and  the  Z£. 

Letting  #  represent  the  length  of  AH, 
show  that  A  B  =  2#,  andB  H=#j/3. 

1 1 .  In    the    right    triangle    A  H  B 


A     *    H          C 

(Fig.  137),  show  that  the  sine,  cosine, 
and  tangent  of  30°,  and  of  60°,  have 
the  values  indicated  in  the  following  table : 


H 

FIG.  137 


Sine 

Cosine 

Tangent 

0 

* 

i    /  — 

i    /—  * 

_ 

vv  5 

3V   0 

60°       

4i/3~ 

i 

/  — 

V  6 

*  Prov 

e  that  -7=  =  $V 

'3- 

Compare  these  values  with  the  results  found  in  Problems  7 
and  8.  A  B 

12.  Draw  an     isosceles    right    triangle 
(ABC,  Fig.  138). 

Prove  that  Z  A  =  /_  B  =45°. 

Letting  x  represent  the  length  of  each  of 
the  equal  sides,  show  that  A  E=x\/2. 

13.  From  Fig.  138,  show  that  the  sine,  cosine,  and  tangent 
of  45°  have  the  values  indicated  in  the  following  table: 


x 


Sine 

Cosine 

Tangent 

4<;0. 

4  i/a* 

4  1/2* 

I 

*  Prove  that—  —  =  4. 1/2.     Then  since  1/2  =  i  .4143  +  ,  —^  or  J 

1/2  |/  2 

.7071  +.    Compare  these  values  with  results  found  in  Problem  9. 


TABLE  OF  SINES,  COSINES,  AND  TANGENTS  OF 
ANGLES  FROM  i°-8g° 


Angle 

Sine 

Cosine 

Tangent 

Angle 

Sine 

Cosine 

Tangent 

l° 

•0175 

.9998 

•OI7S 

46° 

•7193 

.6947 

1-0355 

2 

•°349 

•  9994 

•0349 

47 

•7314 

.6820 

1.0724 

3 

•0523 

.9986 

.0524 

48 

•7431 

.6691 

I  .1106 

4 

.0698 

.9976 

.0699 

49 

•  7547 

.6561 

1.1504 

5 

.0872 

.9962 

.0875 

50 

.7660 

.6428 

1  .1918 

6 

.1045 

•9945 

.1051 

5i 

.7771 

.6293 

1-2349 

7 

.1219 

•9925 

.1228 

52 

.7880 

•6l57 

1.2799 

8 

.1392 

•99°3 

.1405 

53 

.7986 

.6018 

1.3270 

9 

.1564 

•9877 

.1584 

54 

.8090 

•5878 

I-3764 

10 

•1736 

.9848 

•I763 

55 

.8192 

•5736 

I  .4281 

ii 

.1908 

.9816 

.1944 

56 

.8290 

•5592 

1.4826 

12 

.2079 

.9781 

.2126 

57 

•8387 

•5446 

1-5399 

13 

.2250 

•9744 

•2309 

58. 

.8480 

•5299 

i  .  6003 

14 

.2419 

•97°3 

•2493 

59 

•8572 

•5150 

1.6643 

15 

.2588 

•9659 

.2679 

60 

.8660 

.5000 

1.7321 

16 

.2756 

.9613 

.2867 

61 

.8746 

.4848 

i  .  8040 

i? 

.2924 

•9563 

•3°57 

62 

.8829 

•4695 

1.8807 

18 

.3090 

•9511 

•3249 

63 

..8910 

•4540 

i  .9626 

iQ 

•3256 

•945-5 

•3443 

64 

.8988 

•4384 

2.0503 

20 

.3420 

•9397 

.3640 

65 

.9063 

.4226 

2  •  1445 

21 

.3584 

•9336 

•3839 

66 

•9i35 

.4067 

2  .  2460 

22 

•3746 

.9272 

.4040 

67 

.9205 

•39°7 

2-3559 

23 

•39°7 

.9205 

•4245 

68 

.9272 

•3746 

2-4751 

24 

.4067 

•9i35 

•4452 

69 

•9336 

•3584 

2  .  605  I 

25 

.4226 

.9063 

.4663 

70 

•9397 

.3420 

2-7475 

26 

•4384 

.8988 

.4877 

7i 

•9455 

•3256 

2  .9042 

27 

•4540 

.8910 

•5°95 

72 

•9511 

.3090 

3-0777 

28 

•4695 

.8829 

•53i7 

73 

•9563 

.2924 

3.2709 

29 

.4848 

.8746 

•5543 

74 

.9613 

.2756 

3-4874 

30 

.5000 

.8660 

•5774 

75 

•9659 

12588 

3-7321 

31 

•515° 

•8572 

.6009 

76 

•9703 

.2419 

4.OIO8 

32 

•5299 

.8480 

.6249 

77 

•9744 

.2250 

4-33I5 

33 

•5446 

.8387 

.6494 

78 

.9781 

.2079 

4  •  7046 

34 

•5592 

.8290 

•6745 

79 

.9816 

.1908 

5  •  J446 

35 

•5736 

.8192 

.7002 

80 

.9848 

•1736 

5-6713 

36 

.5878 

.8090 

.7265 

81 

.9877 

.1564 

6-3138 

37 

.6018 

.7986 

•7536 

82 

•9903 

.1392 

7-"54 

38 

•6i57 

.7880 

•7813 

83 

•9925 

.1219 

8.1443 

39 

•6293 

.7771 

.8098 

84 

•9945 

.1045 

9-5I44 

40 

.6428 

.7660 

.8391 

85 

.9962 

.0872 

11.4301 

4i 

.6561 

•  7547 

.8693 

86 

.9976 

.0698 

14.3006 

42 

.6691 

•7431 

.9004 

87 

.9986 

•0523 

19.0811 

43 

.6820 

•73J4 

•9325 

88 

•9994 

•0349 

28.6363 

44 

.6947 

•7r93 

•9657 

89 

.9998 

•0175 

57.2900 

45 

•  7071 

.7071 

I  .  OOOO 

io8  Second-Year  Mathematics 

139.  The  preceding  table  gives  the  values,  correct  to  four 
decimal  places,  of  the  sine,  cosine,  and  tangent  of  angles  i° 
to  89°,  inclusive. 

Notice  how  the  sine,  cosine,  and  tangent  change  as  the 
angle  increases  from  i°  to  89°. 

Problems  on  Right  Angles 

In  measuring  angles  needed  for  the  solution  of  problems 
in  science  and  engineering,  the  surveyor's  compass  (Fig.  139) 
and  the  engineer's  transit  (Fig.  140)  are  used. 


FIG.  139 

140.  Imagine  E  H  (Fig.  141)  to  be  rotated  (raised  or 
elevated)  in  the  plane  about  E  to  the  position  E  A.  The  angle 
x,  thus  formed,  is  the  angle  of  elevation  of  A  from  E.  If  E  H 
is  rotated  (lowered  or  depressed)  to  the  position  E  B,  the  angle 
y,  thus  formed,  is  the  angle  of  depression  of  B  from  E. 


Ratio,  Proportion,  Similar  Triangles  109 


FIG.  140 


*^,'y 


H 


FIG.  141 


Second-  Year  Mathematics 


141.  In  the  following  problems  find  the  unknown  parts  of 
the  triangles,  first  from  a  scale  drawing,  and  then  by  the  aid 
of  the  sine,  cosine,  and  tangent  of  the  angles. 

i .  The  rope  of  a  flag-pole  is  stretched  out  so  that  it  touches 
the  ground  at  a  point  20  ft.  from  the  foot  of  the  pole,  and 
makes  an  angle  of  73°  with  the  ground.  Find  the  height  of 
the  flag-pole. 

With  a  ruler  and  protractor,  draw  the 
right  triangle  to  scale  (Fig.  142).  By  measure- 
ment, x  is  found  to  represent  66  ft.  nearly. 

More  accurate  results  can  be  obtained  by 
using  the  tangent  of  Z.A,  thus: 


x  . 

— = tan  7V 
20 


=  3. 2 709, from  the  table  on  p.  107. 


FIG.  142 


Therefore  #  =  20X3.2709=65.418. 

The  angle  A  (Fig.  142)  is  the  angle  of  eleva- 
tion of  the  top  of  the  flag-pole  from  the  point 
A  (see  §  140). 

2.  A  balloon  is  anchored  to  the  ground  by  a  rope  260  ft. 
long,  making  an  angle  A  of  67°  with  the  ground.     If  the  rope- 
line  is  straight,  what  is  the  height  of  the  balloon? 

Use  the  sine  of  angle  A  (§  137). 

3.  A  kite-string  300  ft.  long  is  fastened  to  a 
stake.      The  distance  from  the  stake  to  a  point 
directly  under  the   kite   is   102.6    ft.     Find   the 
height  of  the   kite,  supposing  the   kite-string   to 
be  straight. 

What  is  the  angle  of  elevation  of  the  kite  from 
the  stake  ? 


/J  102.6  C 

FIG.  143 


Draw  the  right  triangle  to  scale  (Fig.  143)  and  measure 
a  and  /.A. 


Ratio,  Proportion,  Similar  Triangles  in 

102.6 

cos  A= —    — =.3420. 
300 

But  cos  70°=  .3420,  from  the  table,  p.  107. 
.*.  A  =  70°,  the  angle  of  elevation  of  the  kite. 

—  =sin  70°=  .9397,  from  p.  107. 
300 

Therefore  a  =  300  X  .  9397  =  281.91. 

The  value  of  a  can  also  be  obtained  from  the  equation 


a  =  l/3oo2  — 102. 62     (See  §  133,  Problem  9.) 

4.  A  vertical  pole,  8  ft.  long,  casts  on  level  ground  a  shadow 
9  ft.  long.     Find  the  angle  of  elevation  of  the  sun. 

Use  the  tangent  ratio. 

5.  The  angle  of  elevation  of  an  aeroplane  at  a  point  A  on 
level   ground,  is  60°.     The  point  C  on  the  ground  directly 
under  the  aeroplane  is  300  yd.  from  A.     Find  the  height  of 
the  aeroplane. 

6.  What  is  the   angle  of  elevation  of  the  top  of  a  hill 
5oo]/3  ft.  high,  at  a  point  in  the  plain  1,000  ft.  from  the  top 
of  the  hill  ? 

7.  What  is  the  angle  of  elevation  of  a  road  that  rises  i  ft. 
in  a  distance  on  the  road,  of  50  feet  ? 

8.  A  road  makes  an  angle  of  6°  with  the  horizontal.     How 
much  does  the  road  rise  in  a  distance  of  100  ft.  along  the 
horizontal  ? 

9.  On  a  tower  is  a  search-light  140  ft.  above  sea-level.     The 
beam  of  light  is  depressed   (lowered)   from  the  horizontal, 
through  an  angle  of  20°,  revealing  a  passing  boat.     How  far 
is  the  boat  from  the  base  of  the  tower  ? 

142.  We  know  that  all  right  triangles  in  which  the  following 
parts  are  equal,  each  to  each,  are  congruent: 
I.     The  two  sides  including  the  right  angle. 


ii2  Second-Year  Mathematics 

II.  A  side  and  one  acute  angle. 

III.  The  hypotenuse  and  one  of  the  other  sides. 

143.  In  other  words,  if  in  a  right  triangle  two  parts  in 
addition  to  the  right  angle  are  given  (at  least  one  being  a  side), 
the  triangle  is  completely  determined,  and  may  be  constructed 
from  these  given  parts. 

144.  The  problems  of  §  141  show  that  when  two  parts  of 
a  right  triangle  are  given  (as  indicated  in  §  142),  the  remaining 
parts  may  be  computed  by  the  methods  of  scale  drawing  or  by 
using  the  sine,  cosine,  and  tangent  of  the  angles. 

Which  method  gives  the  more  accurate  results  ? 

145.  The  branch  of  mathematics  which  shows  how  to 
solve  triangles  by  the  aid  of  the  sine,  cosine,  and  tangent  of 
the  angles  is  called  trigonometry.    The  ratios,  sine,  cosine,  and 
tangent  are  called  trigonometric  ratios.    An  angle  has  three 
other  ratios,  but  they  will  not  be  used  in  this  book. 

PROBLEMS 

146.  Solve  the  following  problems: 

1 .  A  boat  passes  a  tower  on  which  is  a  search-light  1 20  ft. 
above  sea-level.    Find  the  angle  through  which  the  beam  of  light 
must  be  depressed,  from  the  horizontal,  so  that  it  may  shine 
directly  on  the  boat  when  it  is  400  ft.  from  the  base  of  the  tower. 

2.  From  the  top  of  a  cliff  150  ft.  high,  the  angle  of  depres- 
sion of  a  boat  is  25°.     How  far  is  the  boat  from  the  top  of 
the  cliff? 

3.  When  an  aeroplane  is  directly  over  a  town  C  the  angle 
of  depression  of  town  B,  2\  miles  from  C,  is  observed  to  be 
10°.     Find  the  height  of  the  aeroplane. 

4.  From  an  aeroplane,  at  a  height  of  600  ft.,  the  angle  of 
depression  of  another  aeroplane,  at  a  height  of  150  ft.,  is  39°. 
How  far  apart  are  the  two  aeroplanes  ? 


Ratio,  Proportion,  Similar  Triangles  113 

5.  Two   persons,    1,200  ft.    apart,   observe   an   aeroplane 
directly  over  the  straight  line  from  one  to  the  other.     One 
person  finds  the  angle  of  elevation  of  the  aeroplane  to  be  35°; 
the  other,  at  the  same  time,  from  his  position,  finds  it  to  be 
55°.     Find  the  height  of  the  aeroplane. 

6.  On  the  top  of  a  'tower  stands  a  flagstaff.     At  a  point  A 
on  level  ground,  50  ft.  from  the  base  of  the  tower,  the  angle 
of  elevation  of  the  top  of  the  flagstaff  is  35°.     At  the  same 
point  A  the  angle  of  elevation  of  the  top  of  the  tower  is  20°. 
Find  the  length  of  the  flagstaff. 

7.  A  balloon  rises  vertically  from  a  point  C,  on  level  ground. 
At  a  point  A,  200  yd.  from  C,  the  balloon  is  observed  when 
the  angular  elevation  at  A  is  40°.     Six  minutes  later  the  angular 
elevation  at  A  is  60°.     Find  the  altitude  of  the  balloon  at  the 
time  of  the  second  observation,  and  the  number  of  yards  it 
rises  per  second,  supposing  the  motion  to  be  uniform. 

8.  From  a  point  A  on  the  bank  of  a  river  flowing  due  east 
the  angle  of  elevation  of  the  top  of  a  tree  on  the  other  side 
is  45°.     At  a  point  B,  70  yd.  south  of  A,  the  angular  eleva- 
tion is  30°.     Find  the  width  of  the  river. 

9.  At  a  window  20  ft.  from  the  ground,  the  angle  of  de- 
pression of  the  base  of  a  tower  is  15°,  and  the  angle  of  eleva- 
tion of  the  top  of  the  tower  is  37°.     What  is  the  height  of  the 
tower  ? 

10.  A  boy  wishes  to  determine  the  height  H  K  of  a  fac- 
tory chimney.     He  places  a  transit  first  at  B  and  then  at  A 
and  measures  the  angles  x  and  y.    The  transit  is  on  a  tripod, 
3^  ft.  from  the  ground.     A  and  B  are  two  points  in  line  with 
the  chimney  and  50  ft.  apart.     What  is  the  height  of  the  chim- 
ney if  the  ground  is  level  and  if  #=63°  and  ;y =33^°  ?  (Fig.  144.) 

Show  how  the  height  of  the  chimney  could  be  readily  found  by 
measuring  shadow-lengths,  without  using  angles.  One  method  would 
thus  furnish  a  check  on  the  other. 


Second-Year  Mathematics 


In  Fig.  145 


w/z=tan  63°  =  i  .9626 

w 


=tan  33^  =  .6620 


(i) 

(2) 

K 


763C 


FIG.  144 


50      3 

FIG.  145 


I 


(i)  and  (2)  are  simultaneous  equations  in  which  w  and  z 
are  the  unknown  numbers.  To  eliminate  w,  by  substitution, 
we  have,  from  (i) 

7^  =  1.96262     (from  (i))  (3) 

By  substituting  (3)  in  (2), 
i .96262 


=  .6620.     (from  (2))  (4) 

Substitute  this  value  of  z  in  (3), 


50+2 

Find  the  value  of  z  in  (4) . 
thus  obtaining  the  value  of  w. 

11.  A  balloon  is  directly  over  a  straight  road.    The  angles 
of  depression  of  two  buildings  on  the  road  are  34°  and  64°. 
If  the  buildings  are  65  yards  apart,  how  high  is  the  balloon? 

12.  From  a  lighthouse,  situated  on  a  rock,  the  angle  of 
depression  <of  a  ship  is  12°,  and  from  the  top  of  the  rock,  it  is  8°. 

'The  height  of  the  lighthouse  above  the  rock  is  45  ft.     Find 
the  distance  of  the  ship  from  the  rock. 

13.  From  an  aeroplane  the  angles  of  depression  of  the  top 
and  bottom  of  a  flag-pole  55  ft.  high,  are  45°  and  157°,  respec- 
tively.    Find  the  height  of  the  aeroplane. 

14.  Village  B   (Fig.  146)  is  due  north  of  village  C.     An 
army  outpost  is  located  at  a  point  A,  8  miles  due  west  of  C. 


Ratio,  Proportion,  Similar  Triangles 


B  bears  60°  east  of  north  from  A.     An  aeroplane  is  observed 
to  fly  from  C  to  B  in  a  quarter  of 
an   hour.     Find  the  average  hori- 
zontal speed  of  the  aeroplane. 

15.  To  measure  the  width  of  a 
river  flowing  due  east,  a  man  selects 
a  point  A  from  which  a  tree  at  C, 
on  the  other  side  bears  60°  east  of 
north.  He  then  walks  east  from  A 


A 


C 


FIG.  146 


until  he  finds  a  point  B  from  which  C  bears  30°  west  of  north. 
A  B  is  found  to  be  300  yards.     Find  the  width  of  the  river, 

CH  (Fig.  147). 

Show     that     x  =  150,     and 


A 


H      B 


FIG.  147 


1 6.  Two  aeroplanes  start  from 
city  C  at  the  same  time.  Aero- 
plane A  flies  south  at  the  average 
rate  of  15  miles  an  hour.  Aero- 


plane  B  flies  west.  At  the  end  of  £  of  an  hour,  aeroplane  B 
is  observed  to  bear  51^°  west  of  north  from  aeroplane  A.  How 
far  apart  are  the  aeroplanes  at  the  time  of  observation? 
What  is  the  average  speed  of  aeroplane  B  ? 

147.  Problems  on  isosceles  triangles  may  be  solved  by 
using  the  two  right  triangles  into  which  an  altitude  line  from 
the  vertex- angle  divides  it. 

PROBLEMS 

1 .  The  distance  from  a  cannon  to  a  straight  road  is  7  miles. 
If  the  range  of  the  cannon  is  10  miles,  what  length  of  the  road 
is  commanded  by  the  cannon  ? 

Show  that  ABC  (Fig.  148)  is  an  isosceles  triangle,  and  that  A  H 
bisects  B  C.  In  the  right  triangle  A  B  H,  find  the  length  of  B  H. 

2.  The  arms  of  a  pair  of  compasses  are  opened  to  a  dis- 


n6 


Second-  Year  Mathematics 


tance   of   6.25  cm.   between  the   points.      If  the   arms   are 
11.5  cm.  long,  what  angle  do  they  form  ? 

In  the  isosceles  triangle  ABC  (Fig.  149)  draw  the  altitude  A  H. 
Find  the  number  of  degrees  in  x. 


10 


B  H  C 

FIG.  148 

3.  A  pair  of  compasses  is  opened  to  an  angle  of  50°.    What 
is  the  distance  between  the  points  if  the  arms  are  12. 5  cm. 
long? 

Draw  the  altitude  of  the  isosceles  triangle. 

4.  A  cannon  with  a  range  of  1 1  miles  can  shell  a  length  of 
13  miles  of  a  straight  road.     How  far  is  the  cannon  from  the 
road? 

5.  A  clock-pendulum,  20  inches  long,  swings  through  an 
angle  of  6°.     Find  the  length  of  the  straight  line  between  the 
farthest  points  which  the  lower  end  reaches. 

6.  A  clock  pendulum  is  25  in.  long.     Through  what  angle 
does  the  pendulum  swing  if  the  distance  between  the  farthest 
points  which  the  lower  end  reaches  is  6  in.  ? 

7.  Two  firemen  are  playing  a  stream  of  water  on  the  wall 
of  a  burning  building  from  a  fire-hose  which  throws  water 
1 20  ft.     The  distance  on  the  ground  from  the  firemen  to  the 
wall  is  100  ft.     What  is  the  greatest   distance  on  the   wall 
which  can  be  reached  by  the  water  ? 

Relations  of  Trigonometric  Ratios 

148.  Important  relations  between  the   sine,   cosine,   and 
tangent  of  an  angle  can  be  shown  by  simple  formulas. 


Ratio,  Proportion,  Similar  Triangles 


117 


i.  Prove  that  if  A  is  any  acute  angle 
(sin  A)2  +  (cos  A)2  =  i. 

In  Fig.  150  sin  A=a/c 

cos  A=b/c 

Squaring  (i)  and  (2), 

(sin  A)2=a2/c2 
(cos  A)2=&2/c2 

Adding  (3)  and  (4), 


(sin  A)2  +  (cos  A)2  = 

B 


a 


(i) 


(3) 
(4) 

(5) 


A  b  C 

FIG.  150 

In  Fig.  150,  show  that  a2+62=c2. 

Show  that  (5)  may  be  written  (sin  A)3  +  (cos  A)2  =  i.       (6) 
(sin  A)2  is  usually  written  sin2  A:    similarly  (cos  A)2  and 
(tan  A)2  are  written  cos2  A  and  tan2  A. 

2.  In  Fig.  150,  prove  that  sin2  B+cos2  B  =  i. 

3.  Using  the  formula  sin2  #+cos2  x=i,  show  that 

sin  x—V-i  —cos2  x*  (i) 

and  cos  x  =  l/i  —sin2  x  (2) 

4.  The  sine  of  an  angle  x  is  f .     Find  the  cosine  of  x. 
Use  formula  (2),  Problem  3. 

5.  The  cosine  of  an  angle  A  is  §.    Find  the  sine  of  A. 

*  The  double  sign  is  not  used  before  the  root  sign  because  a  negative 
sine  or  cosine  have  no  meaning  under  the  foregoing  definitions  for  acute 
angles  (§138). 


1 1 8  Second- Year  Mathematics 

6.  From  Fig.  150,  show  that 

sin  A 

tan  A= ,  and 

cos  A 

sin  B 

tanB= 

cos  B 

7.  From  the  formulas  of  Problems  3  and  6,  find  cos  y  and 
tan  y,  if  sin  y  =  %. 

Problems  and  Exercises 

149.  The  sine,  cosine,  and  tangent  of  an  angle  are  called 
functions  of  the  angle. 

Find  the  values  of  the  other  two  functions  here  used,  when 

1.  sin  B  =0.5         4.  cosB  =  £j/3       7.  sin  B=o.6 

2.  sinB=J]/3       5.  sin  B  =  Jj/2       8.  cos  B=o.6 

3.  cosB=J  6.  cosB  =  |]/2       9.  cos  B  =0.8 
10.  Find  the  values  of  the  other  functions,  when  tan  B=f. 
From  Problem  6, 

sin  B     , 

C-oiB=f  (I) 

From  Problem  2, 

sin2  B+cos2  B  =  i  (2) 

Equations  (i)  and  (2)  are  to  be  solved  as  simultaneous 

equations  in  the  two  unknowns,  sin  B  and  cos  B.  Sin  B  is 
eliminated  by  substitution,  as  follows: 

From  (i)                             sin  B=f  cos  B  (3) 

Substituting  (3)  in  (2)  T9T  cos2  B+cos2  BN  =  i  (4) 

Clearing  of  fractions  9  cos2  B  +  i6  cos2  B  =  i6  (5) 

25   cos2  B  =  i6  (6) 

cos2  B  =||  (7) 

cos   B=f  (8) 
From  equation  (3)  sin  B  =f  and  cos  B  =f . 


Ratio,  Proportion,  Similar  Triangles  119 

Find  the  values  of  the  other  functions,  when: 

11.  tanB=f  14.  tanB=Jj/3        17.  cosB=J 

12.  tanB  =  i/3       15.  tanB  =  2  18.  sinB=i 

13.  tanB  =  i  16.  tan  6=0.5  T9-  cosB=w 
150.  The  two  fundamental  relations 

sin2  A+cos2  A  =  i 

sin  A 

tanA  =  —   — . 
cos  A 

are  true  for  any  angle.     They  are  therefore  called  identities, 
and  are  sometimes  written 

sin2  A+cos2  A=i  (i) 

_sin  A 

tan  A=—  (2) 

cos  A 

The  symbol,  =  ,  is  read  is,  or  is  identical  with. 
These  fundamental  trigonometric  identities  may  be  used 
in  proving  other  relations  between  the  sine,  cosine,  and  tan- 
gent of  any  angle. 

EXERCISES 
i 


i.  Prove:    i+tan2A= 
From  formula  (2)  tan2  A  = 


Therefore  i  +  tan2  A=  i  + 


COS2  A 
sin2  A 
cos2  A 
sin2  A 


cos2  A 
_cos2  A+sin2  A 
cos2  A 

EE ,  from  formula  (i). 

cos3  A ' 


Prove  the  following  identities: 
sin  A 


2.  cos  A— 


tan  A ' 


cos  A 

3.  tan  A  •  -: -=i. 

sin   A 


i2o  Second-Year  Mathematics 


cos  A    tan  A     sin  A  ' 

5.  cos  A  •  tan  A  •  —  -  r= 

sm  A 

i 

6.  sm  A  •  -  r^cos  A. 

tan  A 


i 


''  tan2  A     sin2  A 

8.  sin  A+cos  A=  (i  +tan  A)  cos  A. 

i          .  i 

— sm  A— cos  A  • 


j.      . T sui  n.  zzn_ua  n  •          — —  . 

sm  A  tan  A 

1    -EEtanA+-    I 


i\j.  T-   -       .  .    —  10,11.   •'i-T^T  IT  • 

cos  A     sin  A  tan  A 

n.  1/1  —sin2  A^sin  A  • . 

tan  A 

12.  tan  A  •  cos  AEEI/I—  cos2  A. 

i      i 

tan2  A~sin2  A ' 


15.  (i+tan2  A)(i— sin2  A)  =  i. 

i  cos  A 

16. 


cos  A  i  —tan  A  cos  A  ' 

i 

17.  --  r—  sm  A  •  tan  A^cos  A. 
cos  A 

18.  tanAliH  --  r) 

\      tan  A/ 


IQ  ___I  __  _ 

LCOS  A    sin  AJL      tan  A 

r  i  __  *  ir  ,   i  i 

LCOS  A    sin  AJL      tan  Aj  ' 

151.  Equations  involving  the  trigonometric  ratios  of  an 
unknown  angle  are  called  trigonometric  equations.  The  value 
of  the  unknown  angle  in  such  equations  may  be  found  by 


Ratio,  Proportion,  Similar  Triangles  121 

algebraic  methods,  aided  by  the    fundamental    identities   in 
§150. 

PROBLEMS 

1.  Find  the  value  of  x  in  cos2#  =  J. 
Taking  the  square  root 

cos  x  =  ±%. 
But  cos  60° =J,  by  Problem  n,  p.  106. 

Therefore,  the  original  equation  is  satisfied  if  60°  be  put 
in  place  of  x;  that  is,  x=6o°  is  a  solution  of  the  equation. 

2.  Find  the  value  of  xm4cos2x=^. 
Dividing  by  4,  cos2#=f. 
Taking  the  square  root      cos  #= 
But                                   cos  3o°  = 
Therefore,  #=30°  is  a  solution  of  the  equation. 

3.  sin  x— — : .     Find  x. 

2  sin  x 

Multiplying  by  sin  x,  sin2^=^ 

sin  x  =  V/\  =  \V  2. 
.'.  #=45°,  by  Problem  13,  p.  106. 

4.  sin  x-\ — : — =4.     Find  re. 

sin  x 

Clearing  of  fractions,  sin2#+i  =|  sin  x 

sin2#— f  sin  x=  —i . 

This  is  a  complete  quadratic  equation  in  which  sin  x  is 
the  unknown  number. 

Solving  by  the  method  of  completing  the  square, 
smax — f  sin  x  +  H  =  -fa 
sin  x— f=±| 
sin  #=f  ±f 
=  2  or    . 


122  Second-Year  Mathematics 

Sin  x  cannot  be  equal  to  2  because  the  sine  of  an  angle  is 
not  greater  than  i  (§138,  Problem  3). 
Therefore  sin  x  =  %. 
Therefore        #=30°. 

5.  tan2*  —  1=  -  .     Find  x. 

cos  x 

It  is  helpful  in  solving  equations  to  express  all  the  functions 
in  terms  of  one  of  them. 

sin2^  i 

By  §150,  (2),  _-!=—  (i) 

cos2*          cos  x 

Clearing  of  fractions      sin2#—  cos2#=cos  x  (2) 

By  §  1  50,  sin2*  =  i  —  cos2x 

.'.  (2)  may  be  written 

1  —  cos2#  —  cos2#=cos  x  (3) 
2  cos2#+cos  x—  i  =o  (4) 

This  is  a  complete  quadratic  equation  in  which  cos  x  is  the 
unknown  number. 

Solving  by  the  method  of  factoring, 

(2  cos  x—  i)(cos  #  +  i)=o  (5) 

2  cos  x—  i  =o,  or  cos  x=$  (6) 
and                                    cos  x  +  i  =o,  or  cos  x=—  i  (7) 

Cos  x  cannot  be  equal  to  —  i  (§138). 

cos  x=% 
And  x=6o°. 

Find  x  in  each  of  the  following: 
2 

6.  tan  x=  —  ^—  . 

tan  x 

T. 

— 


7.  4  sm  x=- 

sm  x 

8.  2  •  tan  x  •  cos  x  =  i. 

g.  3  cos  x  =  2  sin2x. 

Express  sin2*  in  terms  of  the  cosine. 


Ratio,  Proportion,  Similar  Triangles  123 


10.  2  sin  x  =  y"2  •  tan  x. 

11.  tan2*  —  4  tan  #+3=0. 

12.  tan*  —  2-|  --  =o  . 

tan  x 


13.  7  sn 

14.  sin  x—  cos2x=  -]-. 


1  6.  sin  «;-|-sin2^=cos2^. 


17. 

tan* 

18.  2  sin2#+4  cos2*  =  5  cos  x. 


CHAPTER  III 
THE  MEASUREMENT  OF  ANGLES  BY  ARCS  OF  THE  CIRCLE 

152.  Suppose  the  angular  space  of  4R.  A.  about  the  center 
of  a  circle  to  be  divided  into  360  equal  angles.  This  may  be 
done  by  using  the  divisions  on  the  edge  of  a  protractor.  One 
of  the  angles  so  obtained  may  then  be  taken  as  a  unit  of  meas- 
ure for  angles.  It  is  called  a  degree  of  angle.  How  do  the 
sides  of  these  angles  divide  the  circumference?  State  the 
theorem  that  applies. 

Into  how  many  equal  arcs  is  the  circumference  then 
divided  ?  One  of  these  360  equal  arcs  may  be  used  as  a  unit 
of  measure  for  arcs.  It  is  called  a  degree  of  arc.  Thus  a 
degree  of  arc  is  an  arc  intercepted  by  the  sides  of  a  degree  of 
angle. 

If  now  any  central  angle  be  drawn, 
as  ZAB  C  (Fig.  151),  it  will  contain 
a  certain  number,  integral  or  fractional, 
of  these  angular  units,  or  degrees  of 
angle. 

How  many  degrees  of  arc  do  you 
think  there  are  in  the  arc  intercepted 
by  the  sides  of  the  angle  of  Fig.  151  ? 

If  an  angle   contains   15    angular 

units,  how  many  arc-units  are  there  in  the  arc  intercepted  by 
its  sides  ? 

If  a  central  angle  contains  35^  degrees  of  angle,  how  many 
degrees  of  arc  are  there  in  the  intercepted  arc  ? 

If  the  central  angle  contains  m  units  or  degrees  of  angle, 
how  many  degrees  are  there  in  the  intercepted  arc  ?  Why  ? 

124 


Measurement  of  Angles  by  Arcs  of  the  Circle         125 

If  the  central  angle  is  £  of  4  R.  A.  (72  degrees),  what  part 
of  the  circle  is  the  arc  intercepted  by  its  sides  ?  Why  ? 

153.  We  will  now  compare  the  ratio  of  two  central  angles 
with  the  ratio  of  their  intercepted  arcs.  Let  angles  ABC 
and  D  E  F,  in  the  same  circle  or  in  two  equal  circles  (Fig.  152), 
be  any  two  central  angles  such  that  an  angle-unit  x  can  be 
found  that  is  contained  in  each  an  exact  number  of  times. 


FIG.  152 

Assume  it  to  be  contained  in   Z  A  B  C  m  times,  and  in 
Z  D  E  F  n  times. 

Then  ZA  B  C=m  •  Z*  and  ZD  E  F=n  •  Z*. 
w«#_w 
~~     ' 


i.  e.,  the  ratio  of  ZA  B  C  to  ZD  E  F  is  equal  to  the  ratio 
of  their  numerical  measures,  m  and  n. 

The  units  of  angle-measure  into  which  Z  A  B  C  and 
Z  D  E  F  are  divided  will  intercept  m  equal  arcs,  y,  on  A  C,* 
and  n  equal  arcs,  y,  on  D  F.  Why  ? 

Then  A~C=w  •  y,  and  D~F=w  •  y. 
arc  AC     m  •  y    m 

•         _  .  •--  _  ^.   —  .  _  I  ry  \ 

arc  D  F     n  •  y     n  ' 

ZA  B  C    arc  A  C 

Hence  =—  =,  .     Why  ?     State  the  axiom  used. 

Z  D  E  F    arc  D  F 

*  A  C  and  D  F  mean  arc  A  C  and  arc  D  F. 


126  Second-Year  Mathematics 

154.  We  have  proved  that  the  ratio  of  two  commensurable 
central  angles  is  equal  to  the  ratio  of  the  arcs  they  intercept 
on  the  circumference. 

But  the  question  arises,  do  angles  or  other  magnitudes  of 
the  same  kind,  other  than  straight  lines  (§  113),  exist,  for 
which  there  is  no  common  unit  of  measure  ? 

1.  Give  a  common  measure  or  divisor  of: 

n^  ft.  and  4^  ft. 
£  mi.  and  6J  mi. 
3  yd.  and  7$  yd. 
iTV  and  3.25 
6.23  and  i  .6342. 

If  these  number  pairs  are  reduced  to  the  form  of  fractions  having 
a  common  denominator,  the  unit  fraction  having  that  same  denominator 
will  be  a  common  divisor. 

2.  Is  there  a  rational*  common  divisor  or  measure  for: 

3  and  1/^g  ? 
5  and  1/50  ? 
2  and  1/9  ? 

A  simpler  form  for  1/^8  is  31/2,  but  j/2  (the  square  root 
of  2)  cannot  be  expressed  exactly  as  an  integer,  or  a  rational 
fraction.  It  is  found  approximately  to  be  i  .4142  .  .  .  .  ,  and 
the  number  of  decimal  figures  may  be  continued  as  far  as  we 
please,  but  the  exact  value  can  never  be  thus  expressed.  So 
there  is  no  rational  number  that  will  exactly  divide  it. 

Again,  3  and  j/^g  may  be  written  1/9  and  i/^g.  Of  this 
pair,  1/9  seems  to  be  a  common  measure,  as  it  is  contained  in 
1/9  once,  and  in  ]/i8>  1/2  times.  But  1/2  times  as  we  have 
just  seen  is  not  an  exact  integral  or  fractional  number  of  times. 
Hence  3  and  j/ig  have  no  rational  common  measure.  Such 
numbers  are  said  to  be  incommensurable. 

*  A  rational  number  is  a  number  that  can  be  expressed  exactly  as 
an  integer  or  a  fraction  without  the  use  of  the  radical  sign  or  the  equiva- 
lent of  the  radical  sign. 


Measurement  of  Angles  by  Arcs  of  the  Circle         127 

155.  We  shall  see  how  incommensurable  numbers  may  arise. 

1 .  What  is  the  length  of  the  diagonal  A  B  of  a  rectangle 
5  by  7  (Fig.  153)?     (See  §133,  9.) 

Is  there  a  common  divisor  for  5  and  1/74  ?  for  7  and  1/74  ? 

2.  Find  the  length  of  the  diagonal  of  a  square  whose  side 
is  6;   7;  8;  9;   10,   a;  b;  c. 

3.  Is  there  a  common  unit  of  measure  for  the  side  and 
diagonal  of  any  square  ?    What  num- 
ber, incommensurable  with  unity,  or  i, 

does  the  diagonal  always  seem  to  con- 
tain as  a  factor  (see  §-113)  ? 

We  have  seen  that  two  numbers 
and  two  line-segments  may  be  incom- 
mensurable. FlG 

It  is  shown  by  higher  mathematics 

that  the  angles  of  a  scalene  triangle  are  cut  by  the  medians 
into  angles  that  are  incommensurable,  i.  e.,  angles  for  which 
there  is  no  common  angle-unit  of  measure. 

If  then  two  central  angles  are  drawn  equal  to  two  such 
parts  of  an  angle  of  a  scalene  triangle,  would  they  have  the 
same  ratio  as  their  intercepted  arcs?  If  so,  how  can  it  be 
proved,  since  they  have  no  common  unit  of  measure  by  which 
to  express  the  ratio  ? 

Does  the  theorem  proved  for  commensurable  angles  (§  153) 
hold  good  for  incommensurable  angles  ? 

156.  Suppose  ZC  A  D  and  ZE  B  H  (Fig.  154)  to  be  two 
incommensurable  angles.     If  Z  C  A  D,  the  smaller  of  the  two 
angles,  be  used  as  a  unit  of  measure  of  ZE  B  H,  it  will  measure 
it  an  integral  number  of  times,  say  n,  with  a  remainder,  say 
Z  F  B  H.     It  will  intercept  the  same  number  n  of  equal  arcs  in 
E  H  with  remainder  H  F  (see  §  1 53) . 


128 


Second-  Year  Mathematics 


Use  the  remainder,  Z.  F  B  H,  as  a  measure  of  the  first 
measure,  Z  C  A  D.  It  is  contained  in  it  an  integral  number 
of  times,  say  m,  with  a  remainder,  if  the  angles  are  incom- 
mensurable, but  the  remainder  must  be  smaller  than  ZF  B  H; 
but  the  number  of  angles  measured  off  in  •  Z  C  A  D  is  the 
same  as  the  number  of  equal  arcs  of  arc  C  D  intercepted  by  the 
sides  of  ZFBH. 


FIG.  154 

Continue  this  process  indefinitely,  and  the  remainders  be- 
come less  each  time  the  process  is  repeated,  and  the  numerical 
measures  of  the  angles  CAD  and  E  B  H  are  found  to  equal 
the  numerical  measures  of  the  arcs  C  D  and  E  H  to  an  error 
as  small  as  may  be  desired. 

The  ratio  then  of  the  angles  is  equal  to  the  ratio  of  the 
arcs.  For  if  it  be  said  the  ratios  differ  by  any  amount,  d,  we 
could  by  carrying  the  process  sufficiently  far  prove  they  differ 
by  less  than  d  (see  §  121). 

157.  A  rigorous  treatment  of  incommensurable  cases  is 
beyond  the  bounds  of  a  secondary  text.  Hereafter  we  will 
give  proofs  for  commensurable  cases  only,  assuming  the 
theorems  to  be  true  for  incommensurable  magnitudes  also. 

We  are  now  prepared  by  proof  for  commensurable  angles, 
and  discussion  for  incommensurable  angles,  to  recognize  the 
truth  of  the  following  theorem: 


Measurement  of  Angles  by  Arcs  of  the  Circle         129 


PROPOSITION  I 

158.  Theorem:    In  the  same  circle  or  in  equal  circles,  two 
central  angles  have  the  same  ratio  as  the  arcs  intercepted  by 
their  sides. 

Let  ABC  (Fig.  155)  be  any  central  angle.  By  the  pre- 
ceding theorem  it  may  be  compared  with  the  perigon  ABA. 

ZABC    arc    AC        .... 

Then  = ......     Why? 

ZABA    arcAMA 

Expressing  ZA  B  A  and  arc 
A  M  A  by  the  number  of  units 
(degrees  of  angle  and  degrees  of 
arc)  they  contain,  respectively, 
we  have 

ZABC  =  arc  AC 
360  360 

Multiplying  through  by  360, 

ZABC=arc    AC,    in    the 
sense  that  the  number  of  angle  degrees  in  Z  A  B  C  is  the  same 
as  the  number  of  arc  degrees  in  the  arc  A  C. 

As  A  B  C  is  any  central  angle,  the  number  of  degrees  in 
any  central  angle  is  the  same  as  the  number  of  degrees  in  the 
arc  intercepted  by  its  sides.  This  is  more  technically  stated 
thus: 

PROPOSITION  II 

159.  Theorem:     A   central  angle  is  measured  by  the  arc 
intercepted  by  its  sides. 

EXERCISES 

1 60.  Solve  the  following  exercises: 

i.  By   Proposition   II,   using  rule   and  compasses   only, 
divide  a  circle  (a)  into  4  equal  arcs; 
(b)  into  2  equal  arcs. 


FIG.  155 


1 3o 


Second-Year  Mathematics 


2.  Divide  a  circle  into  three  arcs  in  the  ratio  of  i  :  2  :  3. 

3.  Find  an  arc  of  45°;  75°;   105°;   165°;   15°. 

4.  If  a  circle  is  divided  into  4  arcs  in  the  ratio  of  i  :  4  :  6  :  7 
what  is  the  number  of  degrees  in  each  arc  ? 

5.  The  area  of  a  circle  is  616  sq.  in.     How  many  degrees 
are  there  in  an  angle  at  the  center  that  intercepts  an  arc  1 1  in. 
long? 

Assume  area  of  circle  =  ?rR2  and  circumference  =  27rR.     Use  ir=-f. 

6.  In  Fig.  1 56,  A  B  is  a  diameter.     The  number  of  degrees 
in  angle  A  O  C  and  B  O  C  are  represented  by  x2-\-$x  and 
3#2  + 1 2X.     Find  the  values  of  x  and  the  number  of  degrees  in 
arcs  A  C  and  B  C. 

A 

£ 


FIG.  156  FIG.  157 

7.  In  Fig.  157,  ZABC  is  a  right  angle.  ZABD  = 
2X2—  3,  and  ZD  B  C  =  iox2  -15. 

Find  the  values  of  x  and  the  number  of  degrees  in  arcs 
A  D  and  D  C. 

Inscribed  Angles 

161.  Let  us  next  consider  what  relation  a  central  angle 
bears  to  an  angle  intercepting  the  same  arc  but  whose  vertex 
is  on  the  circumference. 

An  angle  whose  sides  are  chords  and  whose  vertex  is  on 
the  circumference  is  called  an  inscribed  angle. 


Measurement  of  Angles  by  Arcs  of  the  Circle         131 


CASE  I.     One  side  of  the  inscribed  angle  is  a  diameter. 

In  Fig.  158,  ABC  is  a  central  angle.     Extend  A  B  to 
meet  the  circumference  as  at  D,  and  join  D  and  C. 

Then  Z  A  D  C  is  an  inscribed  angle 
intercepting  the  same  arc  A  C  as  does 
the  central  angle  ABC. 

Compare  angle  ABC  with  the  sum 
of  Z£>  and  ZC  of  the  triangle  B  D  C 
(§  78,  12).  How  do  Z£>  and  ZC 
compare  in  size  ?  Why  (§  28,  3)  ? 

Then  how  does  Z.D  compare  with 
ZABC? 

What  arc  measures  ZA  B  C? 
Then  what  measures  the  inscribed  angle  ZA  D  C? 

CASE  II.     The  center  of  the  circle  lies  between  the  sides  of 
the  inscribed  angle. 


FIG.  159 

As  in  Fig.  159,  draw  an  inscribed  angle  A  B  C  so  that 
the  center  of  the  circle  lies  between  the  sides  of  the  angle. 
Draw  the  diameter  B  D.  By  Case  I  what  arcs  measure 
ZA  B  D  and  ZD  B  C?  What  then  measures  their  sum,  or 
ZABC? 


132  Second- Year  Mathematics 

CASE  III.  The  center  of  the  circle  lies  outside  of  the  in- 
scribed angle. 

As  in  Fig.  160,  draw  an  inscribed  angle  A  B  C  so  that  the 
center  of  the  circle  lies  outside  of  the  angle.  Draw  the  diameter 
B  D.  By  Case  I,  the  half  of  what  arc  measures  ZA  B  D  ? 
ZC  B  D  ?  What  arc  then  measures  the  angle  ABC,  which 
is  the  difference  between  Z  s  A  B  D  and  C  B  D  ? 

Are  there  any  positions  for  an  inscribed  angle  not  included 
by  the  three  cases  just  considered  ?  We  have  now  proved: 

PROPOSITION  III 

162.  Theorem:     An  inscribed  angle  is  measured  by  one- 
half  the  arc  intercepted  by  its  sides. 

163.  An  angle  is  inscribed  in  a  segment  of  a  circle  when 
the  vertex  is  on  the  arc  of  the  segment  and  its  sides  touch  the 
end-points  of  the  arc. 

EXERCISES 

1 .  Prove  that  all  angles  inscribed  in  the  same  segment  of  a 
circle  are  equal. 

2.  Prove  that  all  angles  inscribed  in  a  semicircle  are  right 
angles  (see  Fig.  161).  A 


3 

C 

FIG.  161  FIG.  162 

3.  Prove  that  all  angles  inscribed  in  a  segment  smaller 
than  a  semicircle  are  greater  than  a  right  angle. 

4.  Prove  that  all  angles  inscribed  in  a  segment  larger  than 
a  semicircle  are  less  than  a  right  angle. 


Measurement  of  Angles  by  Arcs  of  the  Circle         133 


5.  Prove:   If  two  chords  intersect  within  a  circle  and  the 
end-points  of  the  sides  of  two  vertical  angles  be  joined  by  two 
straight  lines,  the  triangles  formed  are  mutually  equiangular 
(see  Fig.  162). 

Two  triangles  are  mutually  equiangular  when  each  angle  of  one 
triangle  is  equal  to  the  corresponding  angle  in  the  other. 

6.  If  two  secants 
meet  without  a  circle, 
the  angle  they  form 
is  measured  by  one- 
half  the  difference  of 
the  arcs  included  be- 
tween them. 

Suggestion:      Show 
(Fig.    163)   that    ZB=  FIG.  163 

ZACD-ZB  AC. 

7.  Prove:    If  two  chords  intersect  within  a  circle,  either 
angle  they  form  is  measured  by  one-half  the  sum  of  the  inter- 
cepted arcs  (see  Fig.  164). 

Prove  Z  A  B  C  is  measured  by  i  (/fC+D~E). 

Draw  EC.  A  B  C  is  an  exterior  angle 
of  triangle  EEC.  To  the  sum  of  what 
two  inscribed  angles  is  it  equal  ?  By  what 
arc  then  is  it  measured?  Give  complete 
proof. 

In  similar  manner  show  by  what  arc 
Z  D  B  A  is  measured. 

8.  Prove:  An  angle  formed  by  a 
tangent  and  a  chord,  is  measured 
by  one-half  the  arc  included  by  its 
sides. 

Suggestion:   Prove  by  Fig.  165  Z  A  B  C  is  measured  by  i  B*C. 
Draw  the  diameter  B  D.      Z  A  B  D  being  a  right  angle  (why  ?)  is 
measured  by  one-half  the  semicircle  BCD. 

Z  C  B  D  is  measured  by ? 

Then  what  measures  Z  A  B  "C  ? 


FIG.  164 


134 


Second-  Year  Mathematics 


9.  Similarly  show  by  what  arc  the  angle  C  B  E  is  measured. 

A_ B £ 


io.  Prove  that  an  angle  formed  by  a  tangent  and  a  secant 
meeting  without  the  circle,  is  measured  by  one-half  the  differ- 
ence of  the  included  arcs. 

Suggestion:  In  Fig.  166  prove  angle  B  is  measured  by  one-half 
(AC-A^D).  ADC  is  an  exterior  angle  of  triangle  A  B  D.  ZB  = 
ZADC-ZBAD.  Why? 


FIG.  i 66 

1 1 .  Prove  that  the  angle  formed  by  two  tangents  to  a  circle 
is  measured  by  one-half  the  difference  of  the  arcs  included  by 
them. 

In  Fig.  167,  join  the  two  points  of  tangency  and  use  angle  ADC 
as  in  Fig.  166. 


Measurement  of  Angles  by  Arcs  of  the  Circle         135 

12.  In  laying  a  switch  on  a  railway  track  a  "frog"  is  used 
at  the  intersection  of  two  rails  to  allow  the  flanges  of  the 
wheels  moving  on  one  rail  to  cross  the  other  rail.  Prove  that 
the  angle  of  the  frog,  a  (Fig.  168),  made  by  the  tangent  to 

3 


FIG.  1 68 

the  curve  and  the  straight  rail  D  E,  is  equal  to  the  central 
angle  O,  of  tfc»  arc  B  F. 

(From  "Real  Applied  Problems,"  School  Science  and  Mathematics.) 

PROPOSITION  IV 

164.  Problem:  Upon  a  given  line-sect*  to  construct  a  seg- 
ment of  a  circle  in  which  an  angle  may  be  inscribed  equal  to  a 
given  angle. 

Given  a  line-sect  b;  and  an  angle  A  (Fig.  169). 

To  construct  upon  b  as  a  chord,  a  segment  of  a  circle  in 
which  an  angle  equal  to  angle  A  may  be  inscribed. 

What  lines  form  an  angle  measured  by  the  same  arc  as  an 
inscribed  angle  (see  Exercise  8,  §  163)  ? 

Draw  a  line-sect  equal  to  b  (see  Fig.  169). 
*  Line-sect  is  the  same  as  line-segment. 


136 


Second-  Year  Mathematics 


FIG.  169 


At  one  end-point  A  construct 
an  angle  equal  to  the  given  angle 
A,  as  E  A  C. 

Then  EA  is  to  be  made  a 
chord,  and  A  C  a  tangent  of  a 
circle. 

It  is  required  to  find  the  center 
of  the  circle. 

On  what  line  from  point  A 
does  it  lie  ? 

On  what  other  line  from  the 
chord  E  A  does  it  lie  ? 

Draw  the  circle  and  the  seg- 
ment required. 

Give  proof  of  the  proposition 
in  full. 


EXERCISES 
165.  Solve  the  following  exercises: 

1.  Draw  any  acute  angle  and  a  line-sect.     Construct  the 
segment  required  by  the  method  of  Proposition  IV.     Inscribe 
angles  in  this  segment,   and  compare  them  with  the  given 
angle  by  using  a  protractor. 

2.  Draw  a  right  angle  and  a  line-sect.     Construct  the  seg- 
ment according  to  Proposition  IV.     Inscribe  an  angle  in  it 
and  test  its  size. 

3.  Draw  an  obtuse  angle  and  a  line-sect.     Construct  the 
segment  of  a  circle  according  to  Proposition  IV.     Test  the 
result. 

4.  What  is  the  size  of  the  segment  compared  with  the 
semicircle  in  each  of  Exercises  i,  2,  and  3  preceding  ? 


Measurement  of  Angles  by  Arcs  of  the  Circle         137 

What  previous  truths  about  inscribed  angles  do  these 
exercises  confirm  (see  Exercises  2,  3,  and  4,  §  163)  ? 

5.  What  name  may  be  given  to  the  chord  formed  by  the 
given  line-sect  in  Exercise  2  ?     State  more  than  one  reason 
for  your  answer. 

6.  On  a  given  line-sect,  construct  a  segment  of  a  circle 
that  shall  contain  an  inscribed  angle  of  45°;  of  22§°;    of  135°. 
Construct  the  angles  with  rule  and  compasses  only. 

7.  On  a  given  line-sect,  construct  a  segment  of  a  circle 
that  shall  contain  an  angle  of  60°;    of  30°;    of  120°;    using 
rule  and  compasses  only.     Test  your  results  with  a  protractor. 

8.  To  draw  a  tangent  to  a  given  circle  from  a  point  with- 
out the  circle. 

Let  C  be  the  center  of  the  given  circle,  and  E  any  given  point 
outside  the  circle.  Connect  C  and  E  with  a  straight  line. 

Draw  a  semicircle  on  the  line-sect  C  E  as  a  diameter. 

At  what  point  will  a  line  from  E  be  tangent  to  0  C  ?  Why  ?  Are 
there  two  such  points?  Why?  See  §  163,  2. 

ALGEBRAIC  EXERCISES 
1 66.  Solve  the  following  exercises  algebraically: 

1.  Tf  the  arc  AB  (Fig.  170)  is  4^—8  degrees,  how  many 
degrees  are   there  in  ZADB? 

In   ZACB?     (C  is  the  center 
of  the  circle.) 

2.  If  the  arc  AB  (Fig.  170) 
is  48°,  find  x  and  the  number  of 
degrees  in  Z  A  C  B  and  Z  A  D  B. 

3.  If  the  arc  A  D  C  (Fig.  171) 
is  190°,  find  the  number  of  de- 
grees in  ZT  A  C  and  Z N  A  C. 

.r  IG. 

4.  If  arc  AMC  (Fig.  171)  is 

represented  by  $x2  +x,  find  the  value  of  x,  arc  A  D  C  being  1 90°. 


138 


Second-Year  Mathematics 


5.  The  arcs  and  angle  being  as  marked  in  Fig.  172,  find 
x  and  y. 


FIG.  173 


6.  Find  x  and  y  (Fig.  173),  the  arcs  and  angle  between  the 
secants  being  as  indicated  in  the  figure. 

7.  When   two   tangents  to  a 
circle    make    an    angle    of    60°, 
into  what  arcs  do  they  divide  the 
circle  ? 

8.  Into    what   arcs    do    two 
tangents  at  right  angles  to  each 
other  divide  the  circle  ? 

9.  Two  tangents  include  two 
•arcs  of   a  circle,  one  of  which 

is  four  times  the  other.     How  many  degrees  are  in  the  angle 
they  form  ? 

10.  Two  angles  of  an  inscribed  triangle  are  82°  and  76°. 
How  many  degrees  are  in  each  of  the  three  arcs  subtended 
by  its  sides  ? 

11.  A  triangle  A  B  C  is  inscribed  in  a  circle,   Z4=57°, 
Z.B=66°.    At  A,  B,  and  C  tangents  are  drawn  forming  a  cir- 
cumscribed triangle  A'B'C'.     Find  the  three  angles,  A',  B', 
and  C' (Fig.  174). 

12.  The  angle  between  twb  secants  intersecting  without  a 


Measurement  of  Angles  by  Arcs  of  the  Circle         139 

circle  is  76°.     One  of  the  intercepted  arcs  is  243°.     Find  the 
other. 

13.  Two  angles  of  a  circumscribed  triangle  are  70°  and  80°. 
Find  the  number  of  degrees  in  each  of  the  three  angles  of  the 
inscribed  triangle  whose  vertices  are  at  the  points  of  tangency. 

14.  The  points  of  tan- 
gency of   a  circumscribed 
quadrilateral     divide     the 
circle  into  arcs  in  the  ratio 
of  7  :  8  :  9  :  12.     Find  the 
angles  of  the  quadrilateral. 

15.  The  vertices  of  an 
inscribed    quadrilateral 
divide  the  circle  into  arcs 
in  the  ratio  of  3  :  4  :  5  :  6. 
Find    the    angles    of    the 
quadrilateral. 

1  6.  Two  tangents  to  a  circle  from  an  outside  point  form  an 
angle  of  70°.  What  part  of  the  circle  is  the  larger  arc  included 
by  the  points  of  tangency  ? 

17.  If  a  race  track  is  constructed  on  the  plan  of  Exercise 
1  6,  the  course  being  laid  out  on  two  tangents  at  an  angle  of 
70°  and  the  arc  of  the  circle  included  by  them,  how  much 
farther  than  his  competitor  must  a  rider  go  whose  path  is 
4  ft.  farther  from  the  rail  ? 

Assume  circumference  =  2irR. 

1  8.  The  angle  between  two  secants  is  30°  (Fig.  175).    The 

6x2  +  2ox  +  30 
—  —  —  --  —  — 


number  of  degrees  in  arc  D  L  is  represented  by 


'    \) 


O-v2   _  fj  sv*  _  T   £ 

in  the  arc  B  C.  by  -  -  -  -  .     Find  x  and  the  number 

*-S 
of  degrees  in  each  of  the  two  arcs. 

Reduce  the  fractions  to  lowest  terms. 


140  Second-Year  Mathematics 

19.  In  Fig.  176,  ZA  E  D  is  60°,  arc  B  C  is  represented  by 

x2  +  i2x-4$ 
;    arc  A  D,  by  -  —  .     Find  the  number 


of  degrees  in  each  of  the  two  arcs. 


FIG.  175 

167.  A  method  of  proof,  often  valuable,  for  an  exercise  or 
a  theorem  that  involves  the  comparison  or  the  measurement 
of  angles  is  that  of  bringing  the  given  figure  into  some  con- 
nection with  the  circle. 

For  example : 

Prove:    The  sum  of  the  three  angles  of  a  triangle  is  two 
right  angles. 

In  Fig.  177,  let  ABC  be  any  triangle;  circumscribe  a 
circle  about  it  (see  §  96, 10).  The  three 
inscribed  angles  are  measured  by  one- 
half  the  sum  of  the  three  arcs  A  B,  B  C, 
and  C  A.  But  the  sum  of  the  three  arcs 
A  B,  B  C,  and  C  A  is  the  entire  circum- 
ference. Therefore  one-half  the  circum- 
ference  or  180°  (the  measure  of  two  right 
angles)  is  the  measure  of  the  sum  of  the 
three  angles  of  the  triangle,  or  AA  + Z.B+ /^C  =  2  R.  A. 
Q.E.D. 

EXERCISES 

1 68.  Verify  the  following  exercises  by  relating  the  given 
figure  to  the  circle : 


Measurement  of  Angles  by  Arcs  of  the  Circle         141 


1.  At  a  given  point  in  a  line  one  and  only  one  perpen- 
dicular to  the  line  can  be  drawn. 

Let  the  two  lines  in  Fig.  1 78  be  J_  at  point  A.     Draw  a  circle  with 
A  as  the  center,  and  prove  any  other  J_  at  A  impossible. 

2.  All    right   angles   are 
equal. 

3.  The  sum  of  the  ad- 
jacent angles  formed  by  one 
straight  line  meeting  another 
is  two  right  angles. 

4.  The  sum  of   all  the 
angles  formed  at  a  point  in 

a  line  on  the  same  side  of  the  FlG  1?8 

line  is  a  straight  angle. 

5.  If  the  sum  of  two  adjacent  angles  is  a  straight  angle, 
their  exterior  sides  form  a  straight  line. 

6.  In  an  isosceles  triangle  the  angles  opposite  the  equal 
sides  are  equal. 

7.  If  two  angles  of  a  triangle  are  equal,  the  sides  opposite 
them  are  equal. 

8.  The  sum  of  the  acute  angles  of  a  right  triangle  is  a 
right  angle. 

9.  An  equilateral  triangle  is  equiangular. 

10.  An  equiangular  triangle  is  equilateral. 

1 1 .  If  two  sides  of  a  triangle  are  unequal  the  angles  opposite 
them  are  unequal,  the  greater  angle  lying  opposite  the  greater  side. 

12.  If  two  angles  of  a  triangle  are  unequal  the  sides  oppo- 
site them .     Complete  the  statement  of  the  proposition, 

and  prove  it. 

13.  That  by  drawing  perpendiculars  from  the  mid-points 
of  two  adjacent  sides,  a  circle  can  be  circumscribed  about  any 
rectangle. 


142  Second-Year  Mathematics 

14.  Using  the  result  of  Exercise  13,  prove  that  the  diagonals 
of  a  square: 

(1)  bisect  each  other; 

(2)  are  perpendicular  to  each  other; 

(3)  are  equal; 

(4)  bisect  the  angles  of  the  square. 

15.  Prove  that  the  diagonals  of  a  rectangle  that  is  not  a 
square : 

(1)  are  equal; 

(2)  bisect  each  other; 

(3)  are  not  perpendicular  to  each  other; 

(4)  do  not  bisect  the  angles  of  the  rectangle. 

HIGHEST  COMMON  FACTOR 

169.  In  §  no  it  was  seen  that  to  find  the  value  of  the  ratio 
of  two  magnitudes,  a  unit  of  measure,  or  a  divisor,  common 
to  the  terms  of  the  ratio,  was  needed.  The  largest  integral 
divisor,  or  the  greatest  common  divisor  (G.C.D.),  of  arithmetical 
or  algebraic  numbers  is  used  in  reducing  ratios,  or  fractions, 
to  lowest  terms,  and  also  to  obtain  the  lowest  common  mul- 
tiple (L.C.M.)  of  the  denominators  of  fractions  to  be  added 
or  subtracted,  and  to  rid  equations  of  fractions. 

A  method  employed  with  arithmetical  numbers  is  as  follows: 

To  find  the  G.C.D.  of  348  and  870. 

The  Method  of  Common  Factors 

Divide  each  of  the  given  numbers  by  the 

2)348,  870         smallest  number  that  will  divide  both,  and 

3)174,  435         repeat  the  process  with  the  successive  quo- 

29)58,  145         dents  as  long  as  they  have  a  common  factor, 

2,     5  or  divisor.     The  product  of  the  divisors  is 

2X3X29  =  174   the  G.C.D.     In  this  example,  174  is  found 

to  be  the  G.C.D.  of  348  and  870. 


Measurement  of  Angles  by  Arcs  of  the  Circle         143 

170.  Another  method  applied  to  line-segments  in  §  112 
is  here  shown  with  arithmetical  numbers. 

The  Method  of  Successive  Division 

Divide  the  larger  by  the  smaller,  and  then  divide  the  divisor 

by  the  remainder,  and  the  last  divisor  by 

348)870(2  the  last  remainder,  repeating  the  process 

696  until  the  division  is  exact.     The  last  divisor 

774)348(2         is  the  G.C.D. 

348  If  the  last  divisor  is  i,  unity  is  the  only 

common  divisor,  and  it  is  of  no  value  in 
reducing  fractions,  because  the  quotients  obtained  by  dividing 
by  i  are  the  same  as  the  original  numbers.  When  the  last 
divisor  is  unity  we  say  the  numbers  are  prime  to  each  other, 
or  that  they  have  no  common  divisor  except  unity. 
For  example,  find  the  G.C.D.  of  73  and  16. 

The  last  divisor  being  i,  we  are  sure 
no   number  larger  than   i    will   exactly 
divide  16  and  73.     This  can  readily  be 
9)l6(*  tested. 

-  Note   the   important   distinction   be- 

tween prime  numbers  and  incommensur- 
-     .  able  numbers  (see  §  113). 

6  171.  The  principle  on  which  the  pro- 

7)2(2       cess  is  based  is:    the  common  divisor  of 
2  any  two  numbers  is  also  a  common  divi- 

sor of  one  of  them  and  of  the  number 
obtained  by  adding  to,  or  subtracting  from,  that  one  a  multiple 
of  the  other. 

Examples 

Find  the  G.C.D.  of: 
(1)174  and  273          (4)  350  and  425 

(2)  174  and  275          (5)  3542  and  5016 

(3)  263  and  765          (6)  3795  and  2865. 


144  Second-Year  Mathematics 

172.  This  method  can  be  applied  to  algebraic  polynomials. 
The  result  is  usually  called  the  Highest  Common  Factor. 

For  example : 


2+  95^+30 
4ox2  +  136^  +  11  2 
—41)—  41^—82 


The  first  term  of  the  first  remainder,  Sx2,  does  not  contain 
$x2  exactly.  So  we  multiply  it  by  5,  for  as  5  is  not  a  factor 
of  either  of  the  given  numbers,  multiplying  a  remainder  by  5 
will  not  introduce  a  common  factor. 

Again,  —41  is  not  a  factor  of  either  of  the  original  num- 
bers, so  removing  it  from  a  remainder  will  not  remove  or  de- 
stroy a  common  factor.  With  these  modifications  the  process 
is  the  same  as  with  arithmetical  numbers.  In  this  example 
the  last  divisor,  #  +  2,  is  the  H.C.F. 

EXERCISES 

173.  Find  the  H.C.F.  of  the  following  polynomials  by  the 
method  of  successive  division: 

As  in  ordinary  division  it  is  convenient  to  arrange  the  given  poly- 
nomials in  the  ascending  or  the  descending  powers  of  the  same  letter, 
if  they  are  not  already  so  arranged. 

1.  3a2+a4  +  i2a  —  16  and  — 

2.  c2  +  2c3  —  yc  —  20  and  — 

3.  4c2+3c-io  and  7c2+4c3  -3^-15 


Measurement  of  Angles  by  Arcs  of  the  Circle         145 


4.  X  —  2X*+X4  and  2X*  —  2X3— 

5.  <;•*  —  13<;2+36  and  c4—  c3  — 

6.  4a3—  4a2  —  50+3  and  ioa2  —  190+6. 

Find  the  H.C.F.  of  the  following  by  either  method: 

7.  x2  —  2ix+2o  and  x2+6x—  7 

8.  i5(w—  ri)3  and  iow2  — 

9.  #2  —  13^+30  and  5t;2 

10.  m*—m—2  andw2+m—  6 

11.  x^—  *  and  xa  — 


When  three  or  more  numbers  are  involved,  find  H.C.F.  of  two,  then 
the  H.C.F.  of  this  result  and  the  next  number,  and  so  on. 

12.  a^—ab2,  2a4  —  2a2b2,  and  a*  —  2a2b+ab2 

13.  c4-^4j  cs+ds,  cs+d*,  and  c2  +  2cd+d2 

14.  I2(m6—n6),  i8(m4—n4),  and  24 

(m3—  mn2—  m2n+n3) 

15.  5^+4063,  7a3  +  28a26  +  28fl62,  and  3  (a4—  4a262) 

16.  x4—  6+x2,  x4  —  io^c2  +  i6,  and  3(x*  +  2X2—  8). 


CHAPTER  IV 

SIMILARITY   AND   PROPORTIONALITY   IN   CIRCLES 
PROPOSITION  I 

174.  Problem:    To  construct  a  square  equivalent  to  a  given 
rectangle. 

Let  a  and  b  be  given  sides  of  a  rectangle. 
To  construct  a  square  equivalent  to  the  rectangle  a  by  b. 
Upon  a  line-segment  EF  equal  to  the  sum,  a+b,  as  a 
diameter,  draw  a  circle  (Fig.  179).     At  the  point  of  division 
C  between  a  and  b  on  the  diameter, 
erect  a  perpendicular  and  extend  it 
to  the  circle  as  at  D. 

Prove  C  D  to  be  the  side  of  the 
required  square. 

Draw  ED    and    F  D.     ZEDF 
is  a  right  angle.     (Why  ?) 
FlG  D  C    is    a     mean    proportional 

between  the  sects*  of  the  hypotenuse 
of  the  right  triangle  A  D  B.     Why?     See  Proposition  XIII, 

p.  97. 

COROLLARY 

175.  Theorem:    A  perpendicular  to  a  diameter  of  a  circle 
at  any  point,  extended  to  the  circumference,  is  a  mean  propor- 
tional between  the  sects  of  the  diameter. 

EXERCISE 

Construct  a  square  equal  in  area  to  the  rectangle  with 
given  sides        c        and    d    . 

*  The  sects  are  the  parts  a  and  b  into  which  the  line-segment  A  B 
is  cut  by  the  foot  C  of  the  perpendicular. 

146 


Similarity  and  Proportionality  in  Circles  147 

Measure  to  some  scale  the  sides  of  the  given  rectangle  and  of  the 
constructed  square,  and  note  whether  the  product  of  the  numbers  of 
the  first  two  measures  is  equal  to  the  square  of  the  third.  If  there  is  a 
difference,  to  what  is  it  probably  due  ? 

Measure  each  several  times  and  take  the  average  of  the  products  of 
the  numbers  expressing  the  measures  of  the  sides  of  the  rectangle,  and 
the  average  of  the  squares  of  the  numbers  expressing  the  measures  of 
the  side  of  the  square.  How  do  the  two  averages  compare  ? 

176.  Draw  a  circle  and  a  chord  A  B  intersecting  a  chord 
C  D  as  at  E  (Fig.  180).     Measure  the  four  sects  from  the 
point  E.     See  whether  E  B  •  E  A=E  C  •  E  D. 

Draw  on  squared  paper  several 
such  figures  with  intersecting  chords 
in  various  positions.  Measure  care- 
fully, and  note  the  approximate 
equality  of  the  products  of  the  sects 
of  each  of  the  two  chords.  To  what 
is  the  difference,  if  any,  probably  due  ? 

177.  If  in  Fig.  i8o,AE.EB  = 
E  C  •  E  D,  then  A  E  and  E  B  may 

be  made  the  means  of  a  proportion  of  which  E  C  and  E  D 
are  the  extremes  (see  §  119).  Draw  A  C  and  B  D,  and  prove 
the  triangles  ACE  and  B  D  E  similar. 

178.  Show  the  same  triangles  similar  by  considering  the 
angles  only  (Fig.  180).     If  the  triangles  are  similar,  the  corre- 
sponding sides  are  proportional: 

CE  :  AE=EB  :  ED. 

Then,  AE-EB  =  CE-ED.  Why?  Note  carefully  what 
sects  make  the  two  equal  products.  These  steps  give  a  method 
of  proof  for  the  following: 

PROPOSITION  II 

1 79.  Theorem :  If  two  chords  of  a  circle  intersect,  the  product 
of  the  sects  of  one  is  equal  to  the  product  of  the  sects  of  the  other. 


148  Second-Year  Mathematics 

Prove  the  theorem. 

Problem:  Given  the  sides  of  a  rectangle,  to  find  the  side 
of  an  equivalent  square  by  using  Proposition  II.' 

In  any  circle  large  enough,  draw  a  chord  equal  to  the  sum  of  two  ad- 
jacent sides  of  the  rectangle.  Draw  a  radius  to  the  point  of  division. 
What  chord  through  this  point  is  bisected  at  this  point  ? 

EXERCISES 

i.  A  chord  of  a  circle  D  C  (Fig.  181)  cuts  the  mid-point 
of  chord  A  B  at  E,  E  D  is  4  inches  longer  than  E  C,  and 
AB  =  i6  inches.     Find  the  lengths 
of  E  D  and  E  C  approximately  to 


2.  The  sects  of  two  intersecting 
chords  are  x+$  and  x— 6  in  the 
one,  x  +  2  and  #  —  5  in  the  other, 
find    x    and    the    length    of    each 
chord. 

3.  The     sects    of     intersecting 
chords  are  as  given  below.     Find  x  and  the  length  of  each 
chord. 

First  Chord  Second  Chord 

(1)  X-4,  X  +  S  #  +  3,  X-4 

(2)  x+z,  x+6  #—4,  #+18 

(3)  x-s,  x-2  x->j,x+i4. 

PROPOSITION  III 

1 80.  Theorem:  If  from  a  point  without  a  circle  two  secants 
be  drawn  to  the  concave  arc,  the  product  of  one  secant  and  its 
external  sect  is  equal  to  the  product  of  the  other  secant  and  its 
external  sect. 

In  Fig.  182,  what  two  triangles  are  similar  because  mutually  equi- 
angular ? 


Similarity  and  Proportionality  in  Circles 


149 


What  proportion  can  then  be  formed  of  the  secants  and  their  exter- 
nal sects  ? 

Give  complete  proof  of  the  theorem. 


FIG.  182 

EXERCISES 

1 .  If  two  adjacent  sides  of  a  rectangle  are  given,  show  how 
other  equivalent  rectangles  may  be  constructed  by  Proposi- 
tion III  without  finding  the  numerical  measures  of  the  sides 
of  the  given  rectangle. 

2.  Given         a        and     b   ,  the  sides  of  a  rectangle. 

By  use  of  Proposition  III  construct  three  different  rectangles 
equivalent  in  area  to  that  with  sides  a  and  b. 

3.  By  use  of  Proposition  III  can  a  square  be  constructed 
equivalent  to  a  given  rectangle  ? 

What  must  be  the  relative 
size  of  the  whole  secant  and 
its  external  sect?  If  it  were 
drawn  so  as  to  make  its 
external  sect  equal  to  the 
whole  secant  by  what  name 
could  the  "secant"  then  be 
called  ? 

181.  Draw  from  a  point  P 


FIG.  183 


without  a  circle,  a  tangent  PD  (Fig.  183),  and  a  secant  P  F, 
cutting  the  circle  at  some  point,  as  C.     Draw  D  C  and  D  F. 


150  .  Second-Year  Mathematics 

What  triangles  contain  as  sides,  the  tangent,  the  secant,  and 
its  external  sect? 

Show  the  triangles  to  be  mutually  equiangular  and  there- 
fore similar. 

Form  a  proportion  making  P  D  the  mean  proportional. 

PROPOSITION  IV 

182.  Theorem:  If  from  a  point  without  a  circle  a  tangent 
and  a  secant  be  drawn,  the  tangent  is  a  mean  proportional  be- 
tween the  entire  secant  (to  the  concave  arc)  and  its  external 
sect.  Prove. 

EXERCISES 

1.  A  tangent  and  a  secant  are  drawn  from  the  same  point 
without  a  circle.     The  secant  measured  to  the  concave  arc  is 
three  times  as  long  as  the  tangent,  and  the  length  of  its  external 
sect  is  10  feet.     Find  the  length  of  the  tangent  and  secant. 

2.  Two  secants  to  the  same  circle  from  an  outside  point 
are  cut  by  the  circle  into  chords  of  the  circle  and  external 
sects  in  the  ratio  of  5  to  3,  and  5  to  i  respectively.     The  first 
secant  is  8  ft.  long.     Find  the  length  of  the  second  secant. 

The  following  problems  relate  to  two  secants  from  an  out- 
side point  as  in  Exercise  2.  Find  the  length  of  the  second 
secant,  and  test  by  Proposition  III. 

Length  of 
First  Secant 
28ft. 
28ft. 
625  ft. 
25  ft. 
36  ft. 

8.  How  do  the  tests  for  the  five  preceding  problems  show 
whether  Proposition  III  is  true  when  the  sects  of  one  secant 
are  incommensurable  with  those  of  the  other  ? 


Ratio  of  Sects  of 

Ratio  of  Sects  of 

First  Secant 

Second  Secant 

3- 

5  : 

:  2 

3 

:  i 

4- 

3  : 

:  i 

5 

:  2 

5- 

4  : 

i 

5 

:  4 

6. 

4  : 

i 

4  : 

:  3 

7- 

7  = 

2 

7  : 

:  3 

Similarity  and  Proportionality  in  Circles 


9.  In  Fig.  184,  A  R  = 


,  EB  = 


\/2,     Find  the  value  of  x,  and  test  by  Proposition  III. 

c 


FIG.  184 
Solution  of  the  equation  of  Exercise  9  : 

V/I4  •    1/2  =  1/28 


•  1/ 


Squaring  each  member: 


Hence  : 
Test: 


x  =  2 

21/2  +  5  =  114  •  12 
21/7  =  1/2  •  1/7  •  1/2 
21/7  =  21/7 

Remember  that  the  square  of  the  square  root  of  a  number,  is  the 
number  itself,  by  definition  of  square  root.  Thus  the  square  of  1/8, 
[or  0/8)a],  is  8;  (Va  +  b)2  is  a  +  b;  (l/c'+d*)2  is  c*  +  d*. 


10.  In    Fig.    185,    A  6  =  1/9-3;, 


BD  = 


I/ac—  3.     Find  x,  and  test  by 
Proposition  IV. 

Note  that  the  square  of  Vg—X 

is  9  —  x;  that   V5C  +  3  •  Vac—  3  = 

V  x3  —  9;   and  that  the  square  of 

—»  is  81  — 


Are  the  tangents  and 
secants  of  Exercises  9  and 
10  commensurable? 


FIG.  185 


152  Second-Year  Mathematics 

11.  In  Fig.  185,  AB  =  2j/#2  — 9;   B  C=4,  DB  = 

Find  x,  and  the  tangent  and  secant.     Test  by  Proposition  IV. 

12.  In    Fig.    184,    A  B  =  ]/;»— 3,  D  B= — = ,  CB  = 

[,    E  B=— — .     Find  x  and  test  by  Proposition  III. 

l/X  —  2 
Note  that  the  product  of  V x  —  3  by  V x—2  is  x  —  ^Vx  +  6.     Also 


13.  In    Fig.    184,    AB  =  2  +  |/^+4,     D  B  =  2  — 
B  E  =  2  +  j/#+5,   B  0  =  3  —  1/^+5.     Find  x  and  each  line- 
sect.     Interpret  the  result. 

Solution  of  Exercise  13: 

(2  +  V  *+4~)  (2  -  1/  ^+4)  =  (2  +  1/  ^c 


Multiplying  out  4  —  #  —  4=6+  I/  #  +  5—  #  —  5 

-V 
Squaring 

Test:      (2  +  / 


The  principal  square  roots  do  not  satisfy  the  equation. 
Using  the  negative  roots,  2  •  2  =  i  •  4, 

4  =  4- 

The  "secant"  A  B  is  thus  shown  to  be  a  tangent. 
The  symbol  -\/  denotes  the  positive  square  root.     When  the  positive 
and  negative  roots  are  desired  the  symbol  ±  i/  is  used. 


14.  In    Fig.     184,     A~B  =  s  + 

B  C=8—  1/^+3  ,  B  £=3  +  1/^+3.     Find  y  and  each  line- 
sect. 

15.  In    Fig.    184,    AB  =  s-v/5,  BC=8+3i/2,  EB  = 
2—/2.     FindDB. 


1  6.  In    Fig.    184,    DB  =  !/3  :  B  C  =  |/6  +  2i/2  :  B  E  = 
2]/6~—  31/2.     Find  A  B. 

17.  In  Fig.  184,  the  secant  A  B  is  the  square  of  the  external 


Similarity  and  Proportionality  in  Circles  153 

sect.     B  C  =  3l/f-2l/f,  BE  =  2l/J+8l/f.     Find    the    se- 
cant A  B. 

Let  x  be  the  external  sect,  then  x2  is  the  secant. 


#3=6  '§-4  •  ^  +  24  •  i-i6 


1  8.  In   Fig.   185,    the   tangent   A  B   is  1/3  +  51/7.     The 
secant  B  C  is  3  +  51/7.     Find  the  external  sect  B  D. 
Let  *  =  B  D. 

19.  In  Fig.  185,  A  B  =  1/21/3  ,    €6=41/3.      Find    the 
sect  B  D. 

20.  In  Fig.  185,  A  B  =  i,  B  0=1/2  +  1/3.     Show  that  D  B 


21.  In  Fig.  185,  AB=\/l/a+l/6,  B  C  =  Vab,  show  that 


22.  InFig.  185,  AE  =  Vx+y.    The  secant  B  C  =  Vx2-y2. 

Find  the  external  sect  D  B  to  be  equal  to •  l/x2—  ya. 

x—y 

PROPOSITION  V 

183.  Problem:  To  draw  a  common  tangent  to  two  circles 
exterior  to  each  other. 

Let  A  and  A'  (Fig.  186)  be  the  centers  of  two  circles  exterior 
to  each  other.  Connect  A  and  A'. 

Divide  this  line  into  sects  proportional  to  the  radii  of  the 
circles  as  at  C  (see  §  125,  2). 

Such  a  point  of  division  is  called  a  center  of  similitude  of  the  two 
circles. 

From  C  draw  a  tangent  to  circle  A'. 

How  can  the  points  of  tangency  be  found  ?     (See  §  165,  8.) 


154 


Second-Year  Mathematics 


Let  D'  be  one  point  of  tangency.     Draw  D'A'.     Draw 
D'C,  and  extend  it  beyond  circle  A. 


FIG.  186 

Draw  A  D  to  line  D'C  and  parallel  to  D'A'. 

If  A  D  can  be  proved  to  be  a  radius  of  circle  A,  by  using 
the  constructions  made  and  the  fact  of  the  similarity  of  the 
triangles,  then  D'D  is  proved  to  be  a  tangent  to  both  circles- 
Why? 

Similarly  prove  E  F  a  tangent  to  both  circles. 

D  D'  and  E  F  (Fig.  186)  are  called  transverse  tangents. 

184.  Prove  C  D'D  and  C  E  F  common  tangents  to  circles 
'  and  A  (Fig.  187). 


C 


FIG.  187 


Similarity  and  Proportionality  in  Circles 


155 


If  A  A'  be  divided  externally  into  sects  proportional  to  the  radii  of 
the  circles,  as  A  D  and  A'D'  (Fig.  187),  the  constructions  and  proofs 
are  similar  to  those  in  Proposition  V.  Give  full  proof. 

C  D'D  and  C  E  F  (Fig.  187)  are  called  direct  tangents. 
The  point  C'  is  called  an  external  center  of  similitude. 

185.  As  the  distance  between  the  centers  of  the  circles  is 
made  to  decrease,  certain  interesting  changes  are  produced  in 
the  common  transverse  and  direct  tangents,  as  shown  in  Figs. 
188  to  192. 


FIG.  188 


Show  how  the  points  C,  D,  D',  E,  and  F  of  Figs.  1 86  and  187 
change  position  in  Figs.  188  and  189. 

G 


When  do  two  or  more  of  such  points  coincide  ? 
In  which  figures  do  they  disappear  ? 


156  Second- Year  Mathematics 

What  lines  in  Figs.  186  and  187  do  the  lines  H  G  in  Figs.  189 
and  191  represent? 


FIG.  190 


FIG.  191 


FIG.  192 


FIG.  193 


Similarity  and  Proportionality  in  Circles 


157 


1 86.  Prove  that  all  lines  from  a  center  of  similitude  of  two 
circles,  are  cut  proportionally  by  the  circle. 

Prove  by  Proposition  V  that  in  Figs.  193  and  194. 
CB     CD     CE     CF     CH 


CB' 

because  each  is  equal  to 


CA' 


"CF    CH' 


F' 


FIG.  194 

To  Draw  Common  Tangents.     Second  Method 
187.  A  second  method  for  drawing  common  tangents,   is 

here  suggested,  not  based  on  similar  figures. 

To  find  by  analysis  the  points  of  tangency,  B  and  A,  of  the 

common  tangent  to  circles  C  and  C'  (see  §  127). 

-8  A 


FIG.  195 

First,  assume  a  common  direct  tangent  B  A  to  be  drawn  as 
in  Fig.  195,  and  C  B  and  C'A,  to  be  radii  to  the  points  of  con- 
tact. They  are  both  _]_  to  B  A  (why?),  and  therefore  ||  to 


158  Second-Year  Mathematics 

each  other  (why?).  If  C'D  is  drawn  ||  to  A  B,  it  completes 
a  rectangle  A  C'D  B  (why?).  Then  D  B=A  C',  and  D  C  = 
B  C—  A  C'.  If  a  circle  be  drawn  with  C  as  center  and  radius 
C  D  (=B  C-A  C'),  C'D  is  a  tangent  to  it  at  D  (why?),  and 
the  radius  of  circle  C  to  point  of  tangency  of  the  common  tan- 
gent passes  through  D. 

Second,  reverse  this  process. 

In  the  larger  given  circle  draw  a  concentric  circle  D  m 
with  radius =R—r.  From  C',  center  of  smaller  circle,  draw 
a  tangent  to  O  D  m,  locating  point  D  (see  §  165,  8). 

Extend  line  C  D  till  it  meets  larger  circle  as  at  B. 

Draw  C'A  ||  C  B. 

Join  B  and  A. 

B  A  is  the  common  direct  tangent  required.     Tell  why. 

i.  In  similar  manner  draw  B'A',  a  second  direct  tangent. 

1 88.  Assume  a  transverse  tangent  B  A  to  be  drawn  as  in 
Fig.  196,  and  the  radii  C'A  and  C  B,  extended,  drawn  to  the 


FIG.  196 

points  of  tangency.  Then  draw  C'D  ||  A  B,  meeting  C  B  as 
at  D.  D  B  must  equal  C'A  and  be  parallel  to  it,  forming  a 
rectangle  as  in  the  preceding  case  (§  187). 


Similarity  and  Proportionality  in  Circles 


159 


Hence  C  D  =  C  B  +  C'A  (or  R+r),  and  C'D  is  tangent  to 
the  circle  with  center  C  and  radius  R+r  (why?),  and  CD 
passes  through  B,  the  point  of  tangency  of  the  common  tan- 
gent on  the  larger  given  circle.  The  radius  C'A  drawn  parallel 
to  C  D  determines  the  point  of  tangency  A  on  the  smaller 
given  circle. 

Then  to  draw  a  common  transverse  tangent  to  two  given 
circle?  C  and  C',  reverse  the  process  just  given. 

Make  the  construction,  and  give  the  proof. 

i.  In  similar  manner  construct  a  second  transverse  tan- 
gent B'A'  (Fig.  196). 

189.  To  express  the  area  of  a  triangle  in  terms  of  the 
sides  and  the  diameter  of  the  circumscribed  circle. 

In  Fig.  197  let  A  B  C  be  any  triangle,  with  sides  a,  b,  and  c  of  known 
length.  Circumscribe  a  circle  about  it  (see  §96,  10). 


FIG.  197 

Draw  a  diameter  from  B,  as  B  F. 
Draw  a  perpendicular  from  B  to  A  C,  as  B  E. 


The  area  of  a  triangle  A  B  C  = 


,  assuming  the  area  of  a  tri- 


angle equals  one-half  the  base  times  the  altitude.  (i) 

To  express  this  area  in  terms  of  the  sides  and  the  diameter,  B  E 

must  be  expressed  in  these  terms. 

Show  AS  B  E  C  and  A  B  F  similar. 


160  Second-Year  Mathematics 


.-.  —  =  £,   (D  =  diameter  B  F.) 
•    a       D 

BE-*  (,) 


Hence  from  (i)  and  (2)  A  B  C=^=r ,  and  we  have  thus  proved  — 


PROPOSITION  VI 

190.  Theorem:  The  area  of  any  triangle  is  equal  to  the 
product  of  the  three  sides  divided  by  twice  the  diameter  of  the 
circumscribed  circle. 

Give  a  complete  proof. 

EXERCISES 

1.  The  three  sides  of  a  triangle  are  14,  8,  and  12.     The 
diameter  of  the  circumscribed  circle  is  14.1.     Find  the  area 
of  the  triangle. 

2.  The  three  sides  of  a  triangle  are  12, 10,  and  8.     The  area 
of  the  triangle  is  39 . 7.     Find  the  diameter  of  the  circumscribed 
circle. 

3.  The  area  of  a  triangle  by  Proposition  VI  is  expressed 

thus:    A=-T=T.     Express  the  diameter  of  the  circumscribed 

2D 

circle  in  terms  of  the  area  and  the  three  sides  of  the  triangle. 

4.  The  three  sides  of  a  triangle  are  8,  12,  and  14  and  the 
area  is  47 . 66.     Find  the  diameter  of  the  circumscribed  circle. 

5.  Using  the  facts  that  the  area  of  a  triangle  is  |&  times  the 
altitude,  and  the  area  of  the  triangle  is  — ,  find  a  formula 

2  \J 

for  the  altitude  to  the  side  b  in  terms  of  the  other  sides  and  the 
diameter  of  the  circumscribed  circle. 

6.  Express  in  a  similar  way  the  altitude  to  side  a;  to  side  c. 

7.  Test  the  formulas  for  the  three  altitudes  of  a  triangle 
whose  sides  are  8,  12,  and  14,  the  area  47.66;  the  diameter  of 


Similarity  and  Proportionality  in  Circles 


161 


the  circumscribed  circle  14.1.     Multiply  each  altitude  by  one- 
half  its  base  to  find  the  area. 

8.  Test  the  formulas  for  the  three  altitudes  of  a  triangle 
with  sides  10,  8,  12  and  the  area  39.7,  and  the  diameter  of  the 
circumscribed  circle  12.09. 

Solution  of  Quadratic  Equations  by  the  Formula 
191.  Two  lights,  Lt  and  L2  (Fig.  198),  are  placed  6  ft. 
apart  and  the  candlepower  of  L2  is  twice  that  of  Lx.    A  screen, 


FIG.  198 

S,  is  slipped  along  a  rod  until  it  is  equally  illuminated  by  both 
lights.     Find  the  distance,  x,  of  the  fainter  light  from  the 


screen. 


It  is 'a  law  of  physics  that  intensities  of  illumination  from 
two  lights,  are  directly  proportional  to  the  candlepowers  of  the 


162  Second-Year  Mathematics 

lights  and  inversely  proportional  to  the  squares  of  the  dis- 
tances from  the  lights  to  the  screen  that  is  illuminated. 

Calling  Ij   and  I2  the  intensities  of  illumination  of  the 
screen  by  Lx  and  L2,  we  have 


I2  x2 

But  as  the  intensities  are  equal  —      =^  =  i 


Hence,  i  = 

2X2 

Or,  x2  —  i2X—  36=0  (A) 

Adding  72  to  both  sides 

x2  —  1  2^+36  =  72 
Or,  #-6  =  ±6i/2 

And  #  =  6±  61/2  =  14.  48  +  ,  or  —  2.48  + 

The  results  mean  there  are  two  places,  one  14.48+  ft.  to 
the  right  of  LT  as  shown  in  the  figure  and  the  other,  between 
L!  and  L2  and  2.48+  ft.  to  the  left  of  L,. 

The  fundamental  equation  of  the  solution  just  given  is 
equation  (A).  This  is  a  particular  example  of  the  general 
type— 

x2+px+q=o  (B) 

The  solution  of  (B)  solves  at  once  all  equations  of  this  form. 

Add  £--q  to  both  sides  of  (B), 
4 

*2     *2 

x*+px+^=^-q 
144 

Or,  (*+>)'  '-*!=« 

\          2/  4 


Whence,  *=--±$l//>2-4?  (i) 


Similarity  and  Proportionality  in  Circles  163 

Designating  one  root  by  #t  and  the  other  root  by  x2, 

P 
*i  =  --+il//>2-4?  (2) 

2 


and  x,  =  --iV-4g  (3) 

Calling  p  the  coefficient  of  the  first  power  of  x  in  the 
quadratic  x*+px+q=o  and  q  the  absolute  term  of  the 
quadratic,  give  descriptive  phrases  for  the  following: 

~ 


Calling  a;,  the  positive  root  of  x3+px-{-q=o,  and  x,  the 
negative  root,  translate  formulas  (2)  and  (3)  into  words. 

Calculate  the  value  of  xI  +x2  and  state  the  result  in  words. 

Calculate  the  value  of  xt  •  x2  and  state  the  result  in  words. 

Calculate  the  value  of  xt  —x2  and  state  the  result  in  words. 

Equation  (B)  is  itself  a  particular  case  of  ax3+bx+c=o 
in  which  a  =  i.  Hence  a  more  general  form  of  the  complete 
quadratic  is  ax3+bx+c=o.  Let  us  now  solve  this  more 
general  form. 

The  General  Quadratic  Formula 

192.  Solve  the  equation  ax3+bx+c=o,  by  completing  the 
square. 

ax*+bx=—  c 

,  bx        c 

x*+—=  — 

a         a 

,  b1  ,  bx     b3       b3      c     b3-4ac 

Add  —  ,  x3-\  ---  --  =  ---  =— 

4a3  a     4<z2     $a3     a        4a3 

,   b 
Extracting  sq.  root,     x-\  —  = 


20 


164  Second-Year  Mathematics 

This  statement  of  the  value  of  x  is  called  the  General 
Quadratic  Formula,  and  is  used  in  solving  complete  quadratic 
equations.  For  every  quadratic  equation  can  be  made  to 
have  the  form  ax2+bx+c=o,  in  which: 

a  represents  the  coefficient  of  x2, 
b  represents  the  coefficient  of  x, 
and  c  is  the  constant  term  when  standing  on  the  left 
side  of  the  equation. 

Consequently  by  solving  ax2+bx+c=o,  we  have  solved  every 
quadratic  equation  in  one  unknown. 

By  this  formula  solve  3X2  +  $x  —  2=0. 

Here  a  =  3,  6  =  5,  andc  =  —  2. 


.  _ 

6  6 

EXERCISES 
193.  Use  the  Quadratic  Formula  in  solving: 

1.  2:V2+5#  +  2=0  5.    2X2+X  =  I$ 

2.  6x2  —  nx+$=o  6.  ax2  +  (b—  a)x  —  b=o 

3.  2r2—  r—  6=0  7.  2ya  +  (4a+b)y=—2ab 

4.  1  2m2  —  i6am—  30*  =o  Solve  for  y. 

Solve  for  m,  also  for  a.  8.   2Z2  —  (2a  +  b)z+ab=o. 

The  sects  of  intersecting  chords  in  a  circle  are  as  follows. 
Solve  for  x  by  the  Formula. 

First  Chord  Second  Chord 

9-  3*-t»  3^-3  2X  +  i,  Sx-4 

10.  2#  +  5,    X  +  I  X  +  2, 

11.  #—5,  2#+3 

12.  2X  +  2,  3^  —  3 

13.  ax+a,  3^—3  ax+^a,  x+i 


Similarity  and  Proportionality  in  Circles 


165 


14.  In  a  circle  of  radius  26  in.,  what  is  the  length  of  a 
chord  perpendicular  to  a  diameter  at  a  point  10  in.  from  the 
center  of  the  circle  ? 

15.  In  a  circle  of  radius  10  in.  the  shortest  distance  from  a 
given  point  on  the  circumference  to  a  given  diameter  is  8  in. 
Find  the  segments  into  which  the  perpendicular  from  the  point 
divides  the  diameter. 


1 6.  Given  a  fragment  of 
a  waterwheel,  to  construct  a 
new  wheel  of  the  same  size. 
D  C  (Fig.  1 99)  is  by  measure- 
ment 4',  A  B  i'.  Compute 
the  diameter. 


FIG.  199 


17.  WW  is  a  wall  with  a  round  corner  of  dimensions  as 
in  Fig.  200,  A  B,  4',  N  E,  9".  It  is  desired  to  find  the  radius 
of  the  circle  of  which  A  N  B  is  a  part,  and  the  location  of 
the  center. 


FIG.  201 


1 8.  To  draw  an  arc  of  a  circle  as  an  "easement"  of  a 
cornice,  tangent  to  the  cornice  at  B  (Fig.  201)  and  passing 
through  A. 


Problems  16,  17,  and  18  are  taken  from  "Real  Applied  Problems" 
in  School  Science  and  Mathematics. 


1 66  Second-Year  Mathematics 

Problems  in  Physics  Leading  to  Complete  Quadratic  Equations 

The  intensity  of  light  on  a  screen  varies  directly  as  the 
candlepower,  and  inversely  as  the  square  of  the  distance  of 
the  source  of  the  light  to  the  screen.  In  the  language  of  pro- 
portion, the  intensity  is  in  direct  ratio  to  the  candlepower,  and  in 
inverse  ratio  to  the  square  of  the  distance  (see  again  §191). 

194.  Solve  the  following  problems  by  the  Quadratic 
Formula. 

1.  An  electric  lamp  placed  before  a  screen  lights  it  as  in- 
tensely as  two  such  lamps,  each  of  the  same  candlepower  as 
the  first,  but  placed  three  feet  farther  from  the  screen.     How 
far  are  the  two  lamps  from  the  screen  ? 

2.  Two  lights  1,000  ft.  apart,  one  of  twice  the  candlepower 
oi  the  other,  appear  equally  bright  when  brought  nearly  into 
line  by  an  approaching  vessel.     What  is  the  distance  of  the 
vessel  from  the  nearer  light  ? 

3.  As  a  boat  approaches  a  harbor  two  range  lights  are 
brought  nearly  into  line.     It  is  known  that  the  nearer  one  is 
of  500  candlepower,  and  the  other  1,500  candlepower,  and 
that  they  are  one-half  mile  apart.     How  far  is  the  boat  from 
the  nearer  light  when  they  are  of  exactly  the  same  intensity  ? 

4.  A  candle  lights  a  screen  as  brightly  as  do  7  equal  candles 
when  placed  10  ft.  farther  from  the  screen.     How  far  are  the 
lights  from  the  screen  ? 

5.  Assuming  the  law  of  heat  to  be  essentially  the  same  as 
that  of  light,  find  how  far  from  a  stove  a  person  is  if  by  moving 
4  ft.  farther  away,  the  intensity  of  heat  is  one-half  as  great  as 
it  was  before. 

6.  How  far  from  a  fire  was  a  man  if  after  moving  7  rods 
nearer  the  fire  the  intensity  of  heat  was  3  times  as  great  as 
before  ? 


Similarity  and  Proportionality  in  Circles  167 

7.  How  far  is  an  object  from  a  light  if  when  moved  toward 
the  light  3  feet,  the  intensity  is  4^  times  as  great  as  before  ? 

8.  An  electric  lamp  lights  a  screen  as  much  as  b  such  lamps 
of  the  same  candlepower  as  the  first,  but  placed  c  ft.  farther 
from  the  screen.     How  far  are  the  b  lamps  from  the  screen  ? 

9.  Use  the  result  of  Problem  8  to  verify  the  results  of 
Problems  i  and  4. 

10.  How  far  from  a  fire  was  a  man  if  after  moving  n  rods 
nearer  the  fire  the  intensity  of  heat  was  b  times  as  great  as 
before  ? 

11.  Use  the  result  of  Problem  10  to  verify  the  result  of 
Problem  5. 

12.  As  a  vessel  approaches  a  harbor  two  range  lights  are 
brought  nearly  into  line.     It  is  known  that  the  nearer  one  is 
of  a  candlepower,  and  the  farther  one  is  of  b  candlepower, 
and  that  they  are  c  miles  apart.     How  far  is  the  vessel  from 
the  nearer  light  when  the  two  are  of  the  same  intensity  ? 

13.  Use  the  result  of  Problem  12  to  verify  the  results  of 
Problems  2  and  3. 


CHAPTER  V 
INEQUALITIES   IN   TRIANGLES    AND    CIRCLES 

195.  Experimental  and  Theoretical  Treatment  of  the 
Axioms  of  Inequality. 

PROBLEMS  AND  EXERCISES 

1.  Construct  triangles  with  sides  as  follows: 

(1)  a=8  6  =  8  c=3 

(2)  a=8  6  =  9  c=3 

(3)  a=8  6  =  10  c=3 

(4)  a=8  6  =  11  c=3 

(5)  0=8  6  =  12  £  =  3 

What  do  you  notice  in  constructing  parts  (4)  and  (5)  ? 
Explain. 

2.  Is  it  possible  to  construct  a  triangle  having  any  three 
line-segments  as  sides  ?    If  the  lengths  of  two  sides  are  given, 
what  condition  must  the  third  side  fulfil?     Calling  the  three 
sides  of  a  triangle  a,  b,  and  c,  express  by  a  formula  a  relation 
which  must  hold  between  them. 

Use  the  sign  >  or  <,  meaning  greater  than,  and  less  than  (see  §198). 

3.  Construct  triangles  with  sides  as  follows: 

(1)  a=8  6=8  c  =  3 

(2)  a=8  6  =  7  c=3 

(3)  0=8  6=6  c=3 

(4)  a  =  8  6  =  5  c=3 

(5)  «=8  6=4  c  =  3 

What  do  you  notice  in  constructing  parts  (4)  and  (5)  ?  Is 
the  condition  asked  for  in  Exercise  2  fulfilled  by  the  above  sets 
of  sides  ?  Explain. 

168 


Inequalities  in  Triangles  and  Circles 


169 


4.  Was  the  condition  arrived  at  in  Exercise  2  sufficient  to 
make  sure  that  a  triangle  can  be  constructed  ?  What  further 
condition  must  be  fulfilled  ? 

Two  of  the  fundamental  truths  on  which  geometry  is 
commonly  based  are  expressed  as  Axioms  8  and  10,  p.  14. 
Re-read  them. 

The  truths  arrived  at  in  Problems  1-4  will  now  be  shown 
to  be  immediate  consequences  of  Axioms  8  and  10. 

PROPOSITION  I 

196.  Theorem:  The  sum  of  any  two  sides  of  a  triangle  is 
greater  than  the  third  side,  and  their  difference  is  less  than  the 
third  side. 

Given  any  triangle  ABC. 
To  prove:      a+b>c;    b+c>a;    c+a>b. 
a—b<c;    b—c<a;    c—a<b. 
a—c<.b. 


Proof:  Consider  the  two  points 
A  and  B  (Fig.  202).  In  what  two 
ways  are  these  points  connected? 
We  have  then  a+b>c.  Why? 

In  a  similar  way  prove  b+c>a 
and  c+a>b. 

Having  proved     a  +  b  >  c 
and  knowing  b=b 


FIG.  202 


subtracting  obtain      a >c—b,  or  c—b<a. 

In  a  similar  way,  prove  the  remaining  five  inequalities. 

PROBLEMS  AND  EXERCISES 

i.  What  obvious  truth  is  used  in  obtaining  the  last  six 
inequalities  from  the  first  three  in  §  196  ? 


170  Second-Year  Mathematics 

To  keep  sharply  in  mind  on  what  fundamental  truths  our 
statements  are  based  and  for  convenient  reference,  this  truth 
is  here  stated  as  — 

AXIOM  13 

The  differences  obtained  by  subtracting  equal  numbers  from 
unequal  numbers  are  unequal  in  the  same  order  as  the  minuends. 

2.  Three  line-segments,  a,  b,  and  c,  being  given,  it  will  be 
possible  to  construct  a  triangle  having  these  line-segments  as 
sides,  if  the  sum  of  every  two  is  greater  than  the  third  line- 
segment;   because,  as  was  seen  in  the  proof  of  Proposition  I, 
if  these  conditions  are  fulfilled,  the  difference  of  every  two 
sides  is  less  than  the  remaining  side. 

3.  Mention  some  sets  of  three  out  of  the  nine  inequalities 
in  §  196,  from  which  the  remaining  six  would  follow. 

4.  State  by  inequalities  the  relations 
which  must  hold  between   the   sides  of 
the  triangle  in  Fig.  203. 

5.  For  what  values  of  x  do  these  rela- 
F                        tions  hold  as  expressed  in  the  following 

inequalities  ? 


Solution:  x+4  +  g>x+i2:  q  +  x+i2>x+4;  x+i2+x+4>g; 
I3>i2;  2i>4;  2x+i6>g; 

2x>-^ 
*>-3* 

6.  For  what  value  of  x  can  the  following  expressions  repre- 
sent the  lengths  of  the  sides  of  a  triangle  ? 

(1)  a=x—  5        b=x-\-'j        c  =  i6 

(2)  a  =  2X+T)         b  =  2X  +  2         C  =  2I 

(3)  a=x+$        b=S—x        c=i 

(4)  a  =  7  b=x-$        c  =  g 

(5)  a  =  2x  6  =  5  c=4*-7 


Inequalities  in  Triangles  and  Circles  171 

197.  Besides  Axiom  13,  p.  170,  the  following  underlie  the 
solution  of  inequalities: 

AXIOMS 

14.  The  sums  obtained  by  adding  equal  numbers  to  unequal 
numbers  are  unequal  in  the  same  order  as  the  unequal  addends: 
e.g.,  5<8 


7<io 

15.  The  products  obtained  by  multiplying  unequal  numbers 
by  positive  equal  numbers  are  unequal  in  the  same  order  as  the 
multiplicands: 

e.g.,   io<i5 

2=2 

-  M 


20<30 

1  6.  The  products  obtained  by  multiplying  unequal  numbers 
by  negative  equal  numbers  are  unequal  in  the  order  opposite  to 
that  of  the  multiplicands: 

e.g.,    i2<    15 


-36>-45 

17.  The  quotients  obtained  by  dividing  unequal  numbers  by 
positive  equal  numbers  are  unequal  in  the  same  order  as  the 
dividends: 

e.g.,   2o<3o 

2=2 

-D 

io<i5 

1  8.  The  quotients  obtained  by  dividing  unequal  numbers 
by  negative  equals  are  unequal  in  the  order  opposite  to  that  of 
the  dividends: 

e-  g-,    5°>4o 

—  2  -  —  2 

-  D 

—  25  <  —  20 


172  Second-Year  Mathematics 

19.  The  differences  obtained  by  subtracting  unequal  numbers 
from  equal  numbers  are  unequal  in  the  order  opposite  to  that  of 
the  subtrahends: 

e.g.,    20  =  20 
5>  3 


198.  The  symbols  for  inequality  are  >,  <,  and  =t=. 
>  means  is  greater  than,  <  means  is  less  than,  and  4=  means 
is  not  equal  to. 

EXERCISES 

1.  Give  some  examples  for  each  of  the  above  axioms. 
Which  of  these  axioms  were  used  in  the  solutions  of  Prob- 
lem 6,  §  196  ? 

2.  Two  sides  of  a  triangle  are  9  and  24  inches.     Between 
what  limits  must  the  third  side  be  ? 

3.  There  is  $50  in  the  treasury  of  a  club.     The  club  wants 
to  buy  furniture  costing  between  $80  and  $90.    How  much 
should  be  raised  at  least  ? 

4.  To  pass  in  a  certain  study  the  final  grade  must  be  above 
60.     The  final  grade  is  obtained  by  taking  the  average  of  the 
quarterly  grade  and  the  examination  grade.     The  quarterly 
grade  being  80,  find  how  much  at  least  the  examination  grade 
should  be  for  passing  ? 

5.  Other  conditions  being  as  in  Exercise  4,  how  much  at 
least  should  the  examination  grade  be  if  the  quarterly  grade 
counts  for  £  in  the  final  mark  ?     For  £  ?     For  £  ?    For  \  ? 

6.  I  live  at  a  distance  of  2\  miles  from  school.     Leaving 
home  at  7 : 30  o'clock,  at  what  rate  must  I  walk  to  get  to  school 
on  time,  if  school  begins  at  8 : 20  ?    At  8:30?    At  8 145?    At 
9  o'clock  ? 

7.  Solve  Problem  6  for  a  pupil  who  lives  i£  miles  from 
school. 


Inequalities  ^n  Triangles  and  Circles 


173 


8.  A  twentieth-century  limited  train  wants  to  make  the 
distance  between  New  York  and  Chicago  (1,000  miles)  in  less 
than  20  hours.     During  the  first  five  hours  it  goes  at  the  rate 
of  45  miles  per  hour.     During  the  next  7  hours  it  goes  at  the 
rate  of  57  miles  per  hour.     How  fast  should  it  go  thereafter 
to  cover  the  distance  within  the  desired  time  ? 

9.  A's  record  average  speed  on  a  2-mile  run  is  6  miles  per 
hour,  and  B's  is  5!  miles.     How  many  feet  can  A  afford  to 
give  B  as  a  handicap  in  a  2-mile  race  ? 

199.  By  means  of  the  truths  discussed  in  the  foregoing 
articles  of  this  chapter,  prove 
the  following  geometric  exer- 
cises and  propositions: 

1.  Prove   that  a   diameter 
of  a  circle  is  longer  than  any 
other  chord  of  that  circle. 

See     Fig.     204;      notice     that 
KO  +  OL  =  MN.     (Why?) 

2.  Prove  that  the  distance  FIG.  204 
between    the  centers    of    two 

intersecting  circles   is  less  than  the   sum    of  the   radii,  but 
greater  than  the  difference. 

PROPOSITION  II 

200.  Theorem:  Any  point  not 
on  the  perpendicular  bisector  of  a 
given  line-segment  is  unequally 
distant  from  the  extremities  of  the 
given  line-segment. 

Given:  AC=CB;  XY.LAB, 
passing  through  C  (Fig.  205). 

Also,  P  lies  outside  X  Y. 

To  prove  P  A=(=P  B. 


174  Second- Ye,ar  Mathematics 

Proof:  Join  the  point  D,  where  PA  and  XY  intersect, 
with  point  B. 

PB<PD+DB     (?) 
DB=DA     (?) 
/.  PB<PD+DA     (?) 
orPB<PA     (?) 
orPB=^PA.     Q.E.D. 

i.  Repeat  the  proof, 

(a)  taking  P  on  the  left  of  X  Y; 

(6)  below  A  B. 

In  §  28,  4,  p.  13,  we  proved: 

Any  point  on  the  perpendicular  bisector  of  a  given  line- 
segment  is  equally  distant  from  the  extremities  of  that  line- 
segment. 

Using  this  proposition  and  Proposition  II,  make  a  state- 
ment concerning  the  location  of  a  point  P,  which  is  known  to 
be  equally  distant  from  the  extremities  of  a  given  line-segment 
as  A  B.  This  statement  may  be  made  as  follows: 

20 1.  Any  point  equally  distant  from  the  end-points  of  a 
line-segment  must  lie  on  the  perpendicular  bisector  of  the  line- 
segment,  and  any  point  on  the  perpendicular  bisector  must  be 
equally  distant  from  the  end-points  of  the  line-segment.     Or  thus, 
the  locus  of  the  points  that  are  equally  distant  from  the  end-points 
of  a  line-segment,  is  the  perpendicular  bisector  of  that  line- 
segment. 

PROPOSITION  III 

202.  Theorem:     If  a  line  contains  two  points  that  are 
equally  distant  from  the  end-points  of  a  given  line-segment,  it 
is  the  perpendicular  bisector  of  the  given  line-segment. 

Given:  (i)  A  line-segment  A  B  and  an  indefinite  line  X  Y 
(Fig.  206); 


Inequalities  in  Triangles  and  Circles  175 

(2)  The  points  P  and  Q  are  on  X  Y; 

(3)  PA=PB;  andQA  =  QB. 

To   prove  the  line  X  Y  is   the  IX 

perpendicular  bisector  of  A  B.  ' 

Proof:  The  perpendicular  bi- 
sector of  A  B  passes  through  Q.  (?) 
See  §201. 


And  also  through  P.     (?)  /\  £ 

The  perpendicular    bisector    of  \Q 

A  B  then  passes  through  P  and  Q. 

But  the  straight  line  X  Y  passes  I 

through  P  and  Q.     (?)  ll 

FIG.  206 
Therefore,  the  straight  line  X  Y 

must  be  the  perpendicular  bisector  of  A  B.     (?) 

203.  In  Axiom  8,  we  recognized  that  the  shortest  distance 
between  two  points  is  measured  along  the  straight  line  joining 

them,  i.  e.,  all  broken  line-segments 
like  A  P  B  and  A  Q  B  (Fig.  207), 
joining  two  points  like  A  and  B 
are  longer  than  a  straight  line- 
segment,  like  A  B,  joining  these 
two  points. 

*  D  It  is  now  to  be  shown  that  of 

FIG.  207  two    such    broken    line-segments, 

one  of  which  lies  entirely  outside 

of  the  other,  the  one  that  is  nearer  to  the  straight  line  is  the 
shorter. 

PROPOSITION  IV 

204.  Theorem:    If  from  a  point  inside  a  triangle,  line- 
segments  are  drawn  to  the  end-points  of  one  side,  the  sum  of 
these  line-segments  is  less  than  the  sum  of  the  other  two  sides. 


176  Second-Year  Matliematics 

Given:   AAB  C,  a  point  P  inside  the  triangle  (Fig.  208). 
To  prove  AP+PC<AB+B  C. 


A  C 

FIG.  208 

Proof:  Prolong  A  P  until  it  intersects  B  C  at  some  point, 
asD. 

We  have  now: 

AP+PD<AB+BD     (?) 
PC<PD+DC     (?) 

adding:        A P+P D+P C< A B+B D+P D+D C     (?) 
subtracting:  A  P+P  C<  A  B+B  D+D  C     (?) 

.'.  AP+PCXAB+BC.     (?) 

i .  Prove  the  same  proposition  by  prolonging  the  line  C  P. 

EXERCISES 

1.  Two  billiard  balls  are  at  A  and  B,  respectively  (Fig.  209). 
To  what  point  on  the  cushion  X  Y  should  the  ball  at  B 

be  directed  that  it  may,  after  reflection,  hit  the  ball  at  A  ? 

Draw  A  A'  perpendicular  to  X  Y  and  so  as  to  make  C  A'= 
A  C.  Draw  A'B  meeting  X  Y  at  some  point  as  P.  Prove 
that  P  is  the  required  point,  assuming  that  the  reflection  is 
such  as  to  make  the  angle  of  reflection  equal  to  the  angle  of 
incidence  (see  Figs.  209  and  210). 

2.  Prove  that  the  distance  from  B  to  A  by  way  of  P  is 
shorter  than  by  way  of  any  other  point  on  X  Y. 

Use  §  201  and  Axiom  8,  p.  14. 


Inequalities  in  Triangles  and  Circles 


177 


3.  A  ray  of  light  reaching  Q  from  P  (Fig.  211),  by  way 
of  the  reflecting  surface  X  Y,  travels  along  the  shortest  dis- 
tance. Determine  at  what  point  the  ray  of  light  should  strike 
X  Y  to  be  reflected  to  Q. 


FIG.  209 


I 


X 


FIG.  210 


Q 


x- 


FIG.  211 


-Y 


4.  Prove  that  light  is  reflected  so  as  to  make  the  angle 
of    reflection   equal    to   the   angle   of  incidence.     See   Exer- 
cise 3. 

5.  Prove  that  the  sum  of  the 
three    line  -  segments    joining    a 
point  inside   of  a  triangle   with 
the    vertices,    is    less    than    the 
perimeter    of    the    triangle    but 
greater   than    its   semf-perimeter 

(see  Fig.  212).  FIG.  212 


B 


i78 


Second-Year  Mathematics 


6.  Determine  between  what  limits  x  must  lie  in  Fig.  213. 
What  values  could  x  have  if  it  were  required  to  be  an  integer  ? 

7.  Prove  that  the  line  joining  one  vertex  of  a  triangle  with 
the  middle  of  the  opposite  side  is  less  than  one-half  of  the  sum 
of  the  remaining  two  sides. 

In    Fig.    214    make    DE  =  BD;  -O 

prove     BE<BC  +  CE    and     prove 
CE  =  BA. 


6 

FIG.  213 

8.  Prove  that  the  sum  of  the  diagonals  of  a  quadrilateral  is 
less  than  the  perimeter,  but  greater  than  the  semi-perimeter. 

9.  Determine  what  integral  numbers  x  may  represent  in 
Fig.  215. 


9 

FIG.  215 


3X+86 
FIG.  216 


10.  Determine  the  integral  values  of  x  from  conditions 
shown  in  Fig.  216,  5#+4  and  4^—31  being  diagonals. 

1 1 .  Determine  for  what  values  of  x  the  following  inequali- 
ties hold.     Give  a  reason  for  each  step. 

(1)  $X  +  2—X/2 

(2)  ^^<6 
'  x-4 

(  ?\       nv- ^i_ 

\3)  2X 


Inequalities  in  Triangles  and  Circles 


179 


205.  The  properties  of  inequalities  may  be  used  to  solve 
problems  in  more  than  one  unknown,  when  the  unknowns  are 
restricted  to  positive  integers. 

PROBLEMS  AND  EXERCISES 
i.  Find  what  positive  integers  will  satisfy  the  equation 


x=i'j  —  2y     (Why?) 

Since  x  and  y  must  be  positive  integers,  we  have 
17  —  2y>o       (?) 


,  and;y>o.     (?) 
Hence,  the  answer  is: 

y=  i>    2>    3.  4,  5>  6>  7,  8. 
*  =  i5»  i3»  "»  9>  7»  5.  3,  I-     (?) 

2.  Determine  what  positive  integers  will  satisfy  the  follow- 
ing equations: 

(1)  3*+;y-i5=o 

(2)  y=42-gx 

(3)  h-S=-3k. 


17 


FIG.  217 

3.  Drawing  the  graph  of  the  equation  #  +  2^  =  17,  solved 
in  Problem  i,  it  becomes  clear  (Fig.  217)  that  in  order  to  have 


180  Second-Year  Mathematics 

both  x  and  y  positive  integers,  we  must  have  y<8%  and  y>o. 
This  agrees  with  the  algebraic  solution. 

Verify  graphically  the  solutions  of  Problem  2. 

4.  Determine,  both  algebraically  and  graphically,  the  posi- 
tive integral  roots  of  the  following  equations: 

(1)  zx-y  =  i8 

(2)  x—  5^+21=0 

(3)  x  +  >jy=-i5. 


5.  A  pipe-system,  100  ft.  long,  is  to  be  built  from  pipes 
5  ft.  and  15  ft.  long.     How  many  pipes  of  each  kind  could 
be  used  ? 

6.  If  20  yards  of  cloth  were  bought  at  one  price,  and  60 
yards  at  another  price,  and  if  altogether  $42.00  were  spent, 
what  was  the  price  of  each  kind,  neither  kind  costing  less 
than  50  cents  a  yard  ? 

7.  If  the  numerator  and  denominator  of  a  fraction  are  both 
increased  by  5,  the  value  becomes  J.     Of  what  fractions  is  this 
possible  ? 

8.  When  some  boys  are  arranged  four  in  a  row,  one  boy  is 
left  out;   if  they  are  arranged  8  in  a  row,  5  boys  are  left  out. 
Determine  how  many  boys  there  may  have  been,  if  there  can 
be  at  most  8  rows. 

9.  Find  what  numbers  leave  a  remainder  2  when  divided 
by  3,  and  also  leave  a  remainder  5  when  divided  by  9. 

10.  A  sum  of  $45  is  to  be  paid  in  five-dollar  bills  and  one- 
dollar  bills.     Determine  how  many  bills  of  either  kind  should 
be  used,  so  that  the  total  number  of  bills  used  should  be  least. 

11.  Determine   the  positive  integral  values  of  x  and  y 
which  will  satisfy  the  equation 


Inequalities  in  Triangles  and  Circles  181 

Solving  (i)  for  x  in  terms  of  y  and  reducing  the  improper 
fraction  to  a  mixed  number: 


~~ 
x  and  y  will  be  integers  if  -    -  is  an  integer. 

i  ~\~y 
Let  —  -=u,  u  being  an  integer. 

2 

Then  i+y=2U  or 

y  =  2U-i  (3) 

Substituting  this  in  (2) 

X  =  $U  +  2  (4) 

We  verify  by  substitution  that  (3)  and  (4)  satisfy  (i)  for  all  values 
of  u. 

Now,  to  find  the  positive  integral  values  of  x  and  y  which 
satisfy  (i),  we  must  find  what  integral  values  of  u  will  satisfy 
the  inequalities: 

2u  —  i  >o   and3W+2>o 
we  find  2u  >  i    and        3^  >  —  2 
or    w>i  and         «>—  §. 
These  values  are:  w=i,  2,  3,  4  ..... 
Substituting  in  (4):  x=$,  8,  n,  14   ..... 
Substituting  in  (5)  :  y  =  1,3,  5,  7   ..... 

12.  Determine  the  positive  integers  that  will  satisfy  the 
following  equations: 

(1)  4*+5;y  =  2i  (4)  50  +  246  =  291 

(2)  3*+8;y  =  25  (5)  7^-13^  =  12 

(3)  9^  —  2^=17  (6)    2/  —  3W  =  2O. 

13.  The  amount  of  $75  is  to  be  paid  out  in  five-dollar  and 
two-dollar  bills.     How  many  bills  of  each  kind  may  be  used  ? 

14.  Determine  the  numbers  which  will  leave  a  remainder 
3  when  divided  by  5,  and  also  when  divided  by  7. 


1 82  Second-Year  Mathematics 

15.  Two  cogwheels  with  170  and  130  cogs,  respectively,  en- 
gage each  other.     At  a  certain  moment  two  marked  cogs  are 
opposite  each   other;    after  how  many  revolutions  oT  each 
wheel  will  the  same  cogs  again  be  opposite  each  other  ? 

1 6.  At  12  o'clock  the  cars  of  two  street-car  companies  leave 
the  same  point.     Cars  of  company  A  leave  the  same  point 
regularly  every  7  minutes  thereafter;    those  of  company  B 
leave  the  same  point  every  9  minutes  thereafter.     At  what 
time  is  the  car  of  company  A,  5  minutes  ahead  of  the, car  of 
company  B  ? 

17.  Under  the  same  conditions  as  in  Problem  16  find  at 
what  time  the  car  of  company  A  is  5  minutes  behind  that  of 
company  B  ?    At  what  times  then  are  they  5  minutes  apart  ? 

1 8.  Other  conditions  being  unchanged,  find  at  what  time 
the  cars  will  be  i  minute  apart  ?    When  will  they  leave  at  the 
same  moment?    When  will  two  cars  of  company  A  come  in 
succession  ? 

19.  Solve  a  problem  like  Problem  16  if  the  cars  start  at 
8  A.  M.,  those  of  company  A  leaving  every  13  minutes  and  those 
of  company  B  every  8  minutes.     When  will  they  be  2  minutes 
apart  ? 

20.  Ask   and   answer   questions   concerning   Problem    19 
similar  to  those  asked  in  Problem  18. 

21.  A  group  of  boys,  numbering  not  more  than  50,  is  ar- 
ranged in  rows  of  4,  leaving  one  over.     If  arranged  in  rows 
of  5,  there  will  be  2  over.     How  many  boys  are  there  ? 

22.  Determine   the  positive   numbers  which   leave   a  re- 
mainder i  if  divided  by  3,  and  leave  a  remainder  7  if  divided 
by  8. 

23.  Into  what  two  parts  should  71  be  divided  so  that  one 
part  may  be  divisible  by  3  and  the  other  part  by  4  ? 


Inequalities  in  Triangles  and  Circles  183 

24.  Gold  alloy  of  93  per  cent,  is  to  be  made  from  90  per 
cent,  and  95  per  cent,  alloy.     How  many  pounds  of  each  kind 
may  be  used  ? 

25.  How  many  pounds  of  the  above  kinds  of  alloy  may  be 
used  to  make  a  new  alloy  of  92  percent.?     Of  92^  percent.? 

26.  John  says  to  William:    "I  divide  my  age  by  5  and 
have  a  remainder  2.     I  divide  it  by  4  and  have  a  remainder  i." 
What  is  John's  age  ? 

27.  "Tell  me  the  remainder  you  get  when  dividing  your 
age  by  5;  also  when  dividing  it  by  4,"  says  a  boy  to  his  father, 
"and  I  will  tell  your  age."     The  answers  being  3  and  2,  re- 
spectively, find  the  father's  age. 

28.  Says  A  to  B:  "If  I  had  7  times  as  many  eggs  as  I  have 
now  and  if  you  had  8  times  as  many  eggs  as  you  have  now, 
and  you  gave  me  one  of  your  eggs,  we  would  have  the  same 
number  of  eggs."     How  many  has  each  ? 

206.  On  p.  14  we  had  Axiom  6,  which  expresses  that  a 
whole  is  equal  to  the  sum  of  all  its  parts.  Connected  with  this 
is  the  following: 

AXIOM  20 

A  whole  is  greater  than  any  of  its  parts* 

PROBLEMS  AND  EXERCISES 

1.  Prove  that  an  exterior  angle  of  a  triangle  is  greater  than 
either  of  the  opposite  interior  angles. 

2.  Prove  that  the  distance  between  the  centers  of  two  cir- 
cles which  lie  entirely  outside  of  each  other,  is  greater  than 
the  sum  of  the  radii  (see  Fig.  218). 

3.  Prove  that  the  distance  between  the  centers  of  two  cir- 
cles, one  of  which  lies  entirely  within  the  other,  is  less  than  the 
difference  of  the  radii  (see  Fig.  219). 

*  This  axiom  does  not  hold  if  we  consider  also  negative  numbers. 


1 84 


Second-  Year  Mathematics 


4.  Draw  a  diagram  representing  a  triangular  field  with 
angles  36°,  72°,  and  72°.  Measure  and  compare  the  sides. 
Which  side  is  smallest? 


FIG.  218 

5.  Draw  triangles  with  angles  as  follows: 

(x)  4=35°  5=65°  C=8o° 

(2)  ,4=20°  5  =  130°          C=3o° 

(3)  4=ioo0          5  =  70°  C  =  io°. 

Measure  the  sides  of  the  triangles  and 
state  in  each  case  where  the  longest 
side  is  situated  with  reference  to  the 
angles,  and  where  the  shortest  side  is 
situated. 

6.  What    connection    appears    to 
exist  between  the  relative  magnitude 
of  the  angles  of   a  triangle   and  the 
relative  magnitude  of  the  sides  ? 
We  shall  now  show  that  the  truth  which  can  be  inferred 

from  the  constructions  in  Problems  4  and  5  is  a  consequence 

of  the  axioms  and  of  propositions  previously  proved  from  the 

axioms. 

PROPOSITION  V 
207.  Theorem:    If  two  angles  of  a  triangle  are  unequal,  the 

sides  opposite  them  are  unequal,  the  greater  side  lying  opposite 

the  greater  angle. 


FIG.  219 


Inequalities  in  Triangles  and  Circles  185 

Given:    A  A  B  C  (Fig.  220). 


To  prove  A  C>A  B. 

Proof:    Through  the  point  B  draw  a  line  B  D,  so  that 
ZDBC=ZC. 

(How  may  this  construction  be  made  ?) 


B' 


FlG.   220 


Then  DB=DC.     Why? 
Further,  A  D  +D  B  >  A  B.     Why  ?  ' 
AOAB.     Q.E.D. 

EXERCISES 

1.  Prove  that  point  D  (Fig.  220)  lies  on  the  perpendicular 
bisector  of  B  C. 

2.  Prove  the  proposition  (used  in  proving  the  above  propo- 
sition), which  says  that  if  a  triangle  has  two  equal  angles  the 
sides  opposite  them  must  be  equal. 

Using  these  last  two  propositions,  we  now  prove  the  con- 
verse of  Proposition  V,  stated  thus.: 

PROPOSITION  VI 

208.  Theorem:  //  two  sides  of  a  triangle  are  unequal,  the 
angles  opposite  them  are  unequal,  the  greater  angle  lying  oppo- 
site the  greater  side. 

Given:    AA  B  C.    A  B<  A  C  (Fig.  221). 

To  prove  ZC<  Z.B. 


1 86 


Second-  Year  Mathematics 


Proof:   Can  ZC=Z-6?     Give  a  reason  for  your  answer. 
Can  Z  C  >  Z.  B  ?     Give  a  reason  for  your  answer. 
How  then  must  the  angles  C  and  B  compare  in  size  ? 


A 


C 


FIG.  221 


EXERCISES 

1.  Two  sides  of  a  triangle  are  15  and  35;  the  angles  oppo- 
site them  are  #+5  and  5^  +  2,  respectively.     How  large  can 
these  angles  be  ? 

2.  Determine  x  from  the  data  given  in  Fig.  222.     Test  to 
see  whether  the  conditions  of  Proposition  VI  are  fulfilled. 

3.  The  sides  of  a  triangle  are 
28,  25,  and  23;  the  angles  opposite 
them  are  125  —  23;,  73  —  2*,  and 
4#  — 18,  respectively.  Determine 
between  what  limits  x  must  lie, 
and  what  values  the  angles  may 
have. 

4.  A  ship  S  is  observed  simultaneously  from  two  forts  at 
A  and  B,  respectively.    As  observed  from  A,  S  bears  50°  N  of 
E  and  B  bears  10°  S  of  E;   as  observed  from  B,  S  bears  40° 
N  of  W,  and  A  bears  10°  N  of  W.     From  which  fort  is  the 
ship  at  the  greater  distance  ?    Find  the  distance  from  either 
fort,  the  distance  from  A  to  B  being  600  feet. 

5.  The  angle  of  elevation  of  a  tower  from  a  point  on  a 
horizontal  plane  185  feet  from  the  base,  is  47°;   from  a  point 


FIG.  222 


Inequalities  in  Triangles  and  Circles  187 

215  ft.  from  the  base  the  angle  of  elevation  is  43°.     Between 
what  two  numbers  does  the  height  of  the  tower  lie  ? 

6.  Prove  that  in  a  right-angled  triangle,  the  hypotenuse  is 
the  longest  side. 

PROPOSITION  VII 

209.  Theorem:    The  perpendicular  from  a  point  to  a  line 
is  shorter  than  any  other  line  connecting  the  point  with  the  line. 

Given  the  point  A  (Fig.  223)  and  the  line  X  Y  of  indefinite 
length ; 

A 


'  / 

K 
\N> 
\ 

> 

v\ 

\           \ 

c 

D      £ 

FIG.  223 

E    F  \ 

x~ 


ABiXY. 

A  C,  A  D,  A  E,  A  F,  oblique  lines. 

To  prove:  A  B<  A  C,  A  D,  A  E,  A  F,  etc. 


Suggestion:  A  B  is  a  side  in  each  of  the  right  triangles  A  B  C,  A  B  D, 
A  B  E,  A  B  F,  etc.     Show  how  the  theorem  follows  from  this. 

EXERCISES 

1.  Using  Fig.  223,  prove  that  A  C=A  F,  if  B  C=B  F,  and 
conversely. 

2.  Using  Fig.  223,  prove  that  A  F  >A  E,  if  B  F  >B  E. 
Show  also  that  angle  A  E  F  >  angle  A  F  E. 

3.  Using    the    same    figure,    prove    that    AF>AD,  if 
BF>BD. 

Apply  Exercises  i  and  2. 


1 88  Second- Year  Mathematics 

PROPOSITION  VIII 

210.  Theorem:  If  two  sides  of  one  triangle  are  equal  to 
two  sides  of  another  triangle,  but  the  angle  included  between  the 
two  sides  in  the  first  is  greater  than  the  angle  included  by  the 
corresponding  sides  in  the  second;  then  the  third  side  in  the  first 
triangle  is  greater  than  the  third  side  in  the  second. 

Given  A's  A  B  C  and  D  E  F  (Fig.  224) : 
AB=DE;  BC=EF; 
To  prove  A  C>D  F. 


\     >^T    D  ~F 

\         X 


FIG.  224 
Proof:  There  are  three  possibilities: 


II. 
III. 

I.  Z.D> Z_A.  In  this  case  place  AD  E  F  on  AA  B  C 
so  that  D  E  falls  on  A  B,  D  on  A,  E  on  B,  and  E  F  on  the 
same  side  of  A  B  as  B  C. 

The  point  F  therefore  falls  below  A  C,  as  at  F'. 

Then  draw  F'C. 

We  have  then:    B  F'=B  C     (?) 
therefore     ZBF'C  =  ZBCF'    (?) 


Inequalities  in  Triangles  and  Circles 


189 


Further 
and 

therefore 
therefore 


ZAF'C  >ZBF'C  (?) 

ZBCF'>ZACF'  (?) 

ZAF'C  >ZACF'  (?) 

A  C>  A  F'  (?) 

AC>DF  (?) 


.    Q.E.D. 


Another  important  truth  has  here  been  used  for  the  first 
time,  namely: 

AXIOM  21 
If  a>b  and  b>c,  then  a>c. 

II.  ZD=  Z4.  Again  place  AD  E  F  on  A  A  B  C  (Fig. 
225),  so  that  D  E  falls  on  A  B,  D  on  A,  E  on  B,  and  E  F  on 
the  same  side  of  A  B  as  B  C. 


Now,  the  point  F  falls  on  A  C,  as  at  F". 

Why  cannot  F  fall  on  A  C  produced  (§  36;  Exercise  3)  ? 

We  have  consequently, 

AF"<AC     (?) 

AC>DF    (?)     Q.E.D. 

III.  Z£><  Z4  (Fig.  226).    In  this  case,  ZF>ZC.     (?) 
Place  AD  E  F  on  AA  B  C  so  that  E  F  falls  on  B  C,  E 
on  B,  and  F  on  C,  and  E  D  on  the  same  side  of  B  C  as  B  A. 
Repeat  the  form  of  proof  given  under  I. 


190 


Second-Year  Mathematics 


i.  Prove  Case  III  of  the  foregoing  proposition  by  placing 
ADEF  on  AABC,  so  that  DE  falls  on  A  B.  (See 
Fig.  227.  Use  Proposition  IV.) 


F 


FIG.  226 


PROPOSITION  IX 

211.  Theorem:  If  two  sides  of  one  triangle  are  equal  to 
two  sides  of  another  triangle,  the  third  side  of  the  -first  triangle 
being  greater  than  the  third  side  of  the  second;  then  the  angle 
opposite  the  third  side  of  the  first  triangle  is  greater  than  the 
angle  opposite  the  third  side  of  the  second  triangle. 


Inequalities  in  Triangles  and  Circles 


191 


Given  A's  P  Q  R  and  X  Y  Z  (Fig.  228): 
PQ=XY;  QR=YZ;  PR>XZ. 
To  prove  Z<2>ZF. 


P  7?     X  z 

FIG.  228 

Suggestions :    What  do  we  know  about  the  triangle  if  Q  =  Y  ? 

Then,  can  Q  =  Y  if  P  R>X  Z,  as  here  given  ? 

What  do  we  know  about  P  R  and  X  Z  if  Q<  Y  ?     Why  ? 

Then,  is  Q<  Y,  if  P  R>X  Z,  as  here  given  ? 

How,  then,  must  angles  Q  and  Y  compare,  if  P  R>X  Z  ? 

Give  full  proof. 

PROPOSITION  X 

212.  Theorem:    Any  point  not  on  the  bisector  of  an  angle 
is   unequally  distant*-  from    the 
sides  of  the  angle. 

Given  ZAB  C  (Fig.  229); 
B  D  bisects  ZAB  C; 
a  point  P  not  on  B  D ; 
PFJLBA;  PExBC. 

To  prove  P  F^P  E.f 

Proof:  As  P  F  or  P  E  must 
intersect  B  D,  assume  that  PF 
intersects  it  at  some  point,  as  G.  Fie.  229 

*  For  distance  of  a  point  to  a  line,  see  §  99,  p.  53. 
fFor  symbol  ^  see  §  198. 


192  Second-Year  Mathematics 

From  G  draw  a  line  G  H  perpendicular  to  B  C.   Draw  H  P. 
Then    PE<PH.     Why? 

and  P  H<  P  G  +  G  H.    Why  ? 
/.  PE<PG  +  GH.    Why? 
AlsoGH  =  GF.    Why? 

/.   PE<PG  +  GF.    Why? 
or    PE<PF,  i.e.,  PE=£PF.     Q.E.D. 
i.  Repeat  the  proof  taking  P  on  the  left  of  B  D. 

PROPOSITION  XI 

213.  Theorem:    Any  point  on  the  bisector  of  an  angle  is 
equally  distant  from  the  sides  of  the  angle. 

Prove.     (See  FYM,  §  294.) 

Using  this  conclusion  and  Proposition  X,  what  can  be 
said  about  the  location  of  a  point  known  to  be  at  equal  dis- 
tances from  the  sides  of  an  angle? 

The  statement  may  be  made  thus: 

PROPOSITION  XII 

214.  Theorem:    Any  point  equally  distant  from  the  sides  of 
a  given  angle  must  lie  on  the  bisector  of  the  angle;   and  any 
point  on  the  bisector  of  a  given  angle  must  be  equally  distant 
from  the  sides  of  the  angle;  or  thus: 

The  locus  of  all  points  equally  distant  from  the  sides  of  a 
given  angle  is  the  bisector  of  that  angle. 

This  is  the  second  example  of  a  locus  met  in  this  chapter. 

EXERCISES 

1.  What  is  the  location  of  all  points  in  a  plane  which  are 
at  a  distance  of  10  ft.  from  a  given  point  P  ?    At  a  distance 
of  4^  ft.  from  a  point  A  ?    At  a  distance  a  from  a  given  point  ? 

2.  What  is  the  locus  of  all  points  in  the  plane  at  equal  dis- 
tances from  two  given  parallel  lines  ? 


Inequalities  in  Triangles  and  Circles  193 

What  is  the  locus  of  all  points  in  space  equally  distant  from 
two  parallel  lines  ? 

3.  What  is  the  locus  of  all  points  having  a  given  distance 
from  a  given  line  ? 

4.  What  is  the  locus  of  all  points  in  space  known  to  have 
a  given  distance  from  a  given  point  ?     Of  all  points  in  space 
known  to  be  at  equal  distances  from  two  given  points  ?     From 
three  given  points  ? 

5.  What  is  the  locus  of  all  points  in  space  having  a  given 
distance  from  a  given  line  ? 

6.  Given  a  triangle  ABC;   it  is  required  to  determine  a 
point  P  that  shall  be  equally  distant  from  the  three  vertices 
of  the  triangle. 

The  point  P  must  be  at  equal  distances  from  A  and  B  (Fig.  230), 
from  B  and  C,  and  from  C  and  A. 

Where  are  all  the  points  gathered 
which  are  at  equal  distances  from 
A  and  B  ? 

On  what  line  then  must  P  lie  ? 

Where  are  all  the  points  collected 
which  are  at  equal  distances  from 
B  and  C  ?  A  /- 

On  what  second  line  then  must  pIG   2,o 

Plie? 

How  then  may  the  point  P  be  determined  ? 

7.  Draw  a  circle  passing  through  the  vertices  of  an  obtuse- 
angle  triangle. 

How  is  the  center  of  the  circle  found  ? 

8.  Draw  a  circle  passing  through  the  vertices  of  an  acute- 
angle  triangle. 

9.  Prove   that   the   perpendicular  bisectors   of   the   three 
sides  of  a  triangle  pass  through  a  common  point. 


194 


Second-  Year  Mathematics 


10.  Given  a  triangle  ABC:    Determine  a  point  that  has 
equal  distances  from  the  three  sides  of  the  triangle  (Fig.  231). 

1 1 .  Inscribe  a  circle  in  an 
acute-angle  triangle. 

If  a  circle  is  inscribed  in  a 
triangle,  the  sides  of  the  triangle 
are  tangent  to  the  circle. 

12.  Inscribe  a  circle  in  an 
obtuse-angle  triangle. 

13.  Prove  that  thebisectors 


FIG.  231 


C 


of  the  three  angles  of  a  triangle 
pass  through  a  common  point. 
14.  Prove  that  the  three  altitude  lines  of  a  triangle  drawn 
from  the  vertices  pass  through  a  common  point. 

Through  each  vertex  of  triangle  ABC  draw  a  line  parallel  to  the 
opposite  side  forming  a  triangle,  as  D  E  F  (Fig.  232). 

Prove   that   the   altitude    lines  of   A  B  C   are   then   perpendicular 
bisectors  of  the  sides  of  A^  E  F.     Then  use  Exercise  9. 


FIG.  232 

The  propositions  and  exercises  which  have  now  been 
proved  enable  us  to  prove  some  new  propositions  concerning 
inequality  of  lines  and  angles  in  circles. 


Inequalities  in  Triangles  and  Circles 


195 


PROPOSITION  XIII 

215.  Theorem:  In  the  same  circle  or  in  equal  circles,  the 
arcs  subtended  by  unequal  chords  are  unequal  in  the  same  order 
as  the  chords;  and,  conversely,  chords  subtending  unequal  arcs 
are  unequal  in  the  same  order  as  the  arcs. 

Given  circle  A=circle  B  (Fig.  233): 
Chord  C  D  >chord  E  F. 
To  prove  arc  C  D  >  arc  E  F. 


FIG.  233 

Proof:  Draw  radii  A  C,  A  D,  B  E,  and  B  F. 

Using  Proposition  IX,  show  that  ZCAD>ZEBF. 

Place  Q  B*  on  0  A,  so  that  E  B  falls  on  A  C,  E  on  C,  and 
B  on  A,  and  F  on  the  same  side  of  C  as  D. 

Then  B  F  must  come  between  A  D  and  A  C,  as  in  position 
A  F'.     ( ?) 

Hence  E  F  comes  in  the  position  C  F',  and  F'  falls  on  the 
circumference  between  C  and  D. 

Then  arc  C  F'< arc  C  D    (?) 
also  arc  C  F'  =  arc  E  F     (?) 
.'.  arc  E  F<  arc  C  D      ( ?)     Q.E.D. 

*  The  symbol  OB  means  the  circle  whose  center  is  at  B. 


196  Second-Year  Mathematics 

Converse.     Given  O  A  =  0  B  (Fig.  233) : 
Arc  CD>arcEF. 

To  prove  chord  C  D  >  chord  E  F. 

Proof:  Draw  radii  AC,  AD,  B  F,  and  B  E,  and  place 
O  B  on  O  A  so  that  E  B  falls  on  C  A,  as  above. 

Since  arc  C  D>arc  E  F,  the  point  F  will  fall  between  C 
and  D,  as  at  F',  and  the  line  B  F  will  come  on  the  same  side 
of  A  D  as  A  C,  as  in  position  A  F'. 

Then  we  have:    ZCAF'<ZCAD     (?) 

also  ZCAF'=ZEBF     (?) 

.-.  ZCAD  >ZEBF     (?) 

Finally,  use  Proposition  VIII  to  show  that  chord  C  D  > 
chord  E  F.  Q.E.D. 

PROPOSITION  XIV 

216.  Theorem:  In  the  same  circle  or  in  equal  circles,  un- 
equal chords  are  unequally  distant  from  the  center  of  the  circle, 
the  shorter  chord  lying  at  the  greater  distance;  and,  conversely, 
chords  unequally  distant  from  the  center  are  unequal,  the  chord 
at  the  greater  distance  being  the  shorter  chord. 

Given  ©P  =  oQ  (Fig.  234): 
Chord  A  B  > chord  D  E. 
To  prove  P  P'<  Q  Q'. 

Proof: 

PP'JLAB, 

QQ'_LDE. 

Place  0  Q  on  0  P,  so  that  Q  falls  on  P,  D  on  B,  and 
chord  D  E  in  the  position  B  C;  then  Q'  will  fall  on  Q",  and 
PQ"_LBC.  Why? 

DrawP'Q". 


Inequalities  in  Triangles  and  Circles 

Then:     P'B>Q"B.     (?) 

Hence  Z B  Q"P' > Z B  P'Q"     (?) 


197 


Observe  that  P  P'J_A  B,  P  Q"J_B  C,  and  use  Axiom  19  (p.  172). 


FIG.  234 

Hence    P  P'  <  P  Q"     ( ?) 
also  since     PQ"=QQ'     (?) 

PP'  <QQ'     (?)     Q.E.D. 

Converse.     Given  (Fig.  234) 

0  P  =  0  Q;     PP'J_AB;     QQ'_LDE; 
P  P'<  Q  Q' 

To  prove  AB>DE. 

Suggestion:    Proceed  with  the  steps  of  the  foregoing  demonstration 
in  the  opposite  order. 

EXERCISES 

1.  Prove  that  a  side  of  a  regular  inscribed  decagon  is  less 
than  a  side  of  a  regular  inscribed  pentagon,  inscribed  in  the 
same  circle,  but  that  the  side  of  the  decagon  is  greater  than 
half  the  side  of  the  regular  pentagon. 

2.  Show  that  the  greater  the  number  of  sides  of  a  regular 
inscribed  polygon,  the  less  is  the  length  of  one  of  its  sides. 


198  Second-Year  Mathematics 

3.  Prove  that  the  distance  from  the  center  of  a  circle  to 
a  side  of  a  regular  inscribed  polygon  is  greater,  the  greater  the 
number  of  sides  of  the  polygon. 

4.  The  length  of  the  chords  AB  and  B  C  being  6#  — 14 
and  4^  +  20,  respectively  (Fig.  235),  and  the  lines  P  P'  and 
P  P'7  being  16  and  10,  determine  x  and  the  chords. 


We  have:   P'B  =  3^-7  (?) 
P"B  = 


Then:  (3#-7)2  +  i6*  =  Pl     (?)     (see  Problem  9,  p.  101). 

and  (2#  +  io)2  +  io2  =  PlB2  (?) 

=  (2*  +  10)2 


or  :  gx2  —  42X  +  49  +  256  =  ^x2  +  403;  +100+100 

5^  —  82^+105  =o. 

82±T/82*-4  •  5  •  105    . 
•'     *  =  —  Ip  -  (see  §  192  (C),  p.  163). 

82  ±68 

x  =  -  =  i15)  or  if 
10 

Then  AB  =  y6,  or  -5! 
CB=8o,  or  25! 

How  is  the  truth  of  Proposition  XIV  illustrated  by  these 
answers  ? 

5.  The  length  of  the  lines  A  B  and  B  C,  P  P',  and  P  P" 
(Fig.  235)  being  denoted  by  1I}  12,  d1}  and  d2,  respectively, 


Inequalities  in  Triangles  and  Circles 


199 


determine  the  unknown  number  in  each  of  the  following  cases. 
In  every  case  test  by  Proposition  XIV. 


/, 

/, 

d, 

rf, 

(l).. 

2a  —  7 

43  —  14 

2 

i 

(2).., 

6 

12 

U-\-  II 

3M  +  4 

(3).. 

x+* 

*+5 

6 

4 

(4).. 

4/+  14 

IO/  —  2 

6 

7 

217.  The  formula  for  the  solution  of  the  general  quadratic 
equation: 

ax2+bx+c=o, 


30 


(Compare  §  192  (C).) 


Calling  the  first  root  x^  and  the  second  root  x2,  we  have: 


By  addition,  we  find: 


-2b     -b  b 

xI+x,= = — or  — 

2<z        a  a 


By  multiplication: 


X-.X-,  — 


b2  —  (b3— 


(2) 

(3) 


(4) 


By  means  of  Formulas  (3)  and  (4)  the  sum  of  the  roots  and 
the' product  of  the  roots  of  a  quadratic  equation  may  be  deter- 
mined without  solving  the  equation. 


2OO  Second-Year  Mathematics 

Having  given  the  equation 

5#2 
we  determine 


\      12 
)  =  —  > 
/      5 


b         /  —  i  2\ 

the  sum  of  the  roots. 
a 


xtxa=-= -,    the  product  of  the  roots. 

v) 

Knowing  the  sum  and  the  product  of  the  roots,  we  can 
determine  something  about  the  sign  of  each  of  the  roots,  viz.: 

I.  If  xjx2  is  positive,  the  roots  must  have  like  signs;  i.  e., 
they  are  both  positive,  or  both  negative.     Why  ? 

And  if.  Xi+xa  also  is  positive,  each  of  the  roots  is  positive; 
but  if  X-L  +x2  is  negative,  each  of  the  roots  is  negative.  Why  ? 

II.  If  x^Xz  is  negative,  the  roots  must  have  unlike  signs; 
i.  e.,  one  is  positive  and  the  other  one  is  negative.     Why? 

And  if  Xi+xa  is  positive,  the  positive  root  is  greater  in 
numerical  value;  but  if  xI+xa  is  negative,  the  negative  root 
is  greater  in  numerical  value.  Why  ? 

EXERCISES 

218.  By  means  of  the  above  discussion,  determine  in  each 
of  the  following  equations  the  sum  and  the  product  of  the 
roots,  without  solving  the  equations  (but  clearing  of  fractions 
when  necessary).  Discuss  also  the  signs  of  the  roots: 

i.  sxz  +  i2X— 18=0 

3.    4#2  —  I2X  —  8=O 

4- 


T.X  —  2       4  —  2X 

5.  ox = — 

°     V  *  —X 


Inequalities  in  Triangles  and  Circles  201 

219.  From  the  formula  for  the  solution  of  the  general  quad- 
ratic equation,  may  be  derived  a  way  of  testing  whether  the 
answers  of  a  given  quadratic  equation  have  a  meaning  or  not 
in  our  ordinary  number-system,  which  is  called  the  system  of 
real  numbers. 

Thus,  if  b2—  4ac  is  negative,  the  expressions  for  the  roots 
of  the  equation  contain  the  square  root  of  a  negative  number. 
But  the  square  root  of  a  negative  number  does  not  exist  among 
the  numbers  which  we  have  thus  far  dealt  with;  i.  e.,  it  is 
not  a  real  number.  Why  not  ? 

So  that  we  have  the  following  conclusions: 

If  b2—  4oc<o,  the  roots  are  not  real. 

If  b3—  4ac>o,  the  equation  has  two  unequal  roots. 

What  can  be  said  about  the  roots  of  the  equation  in  case 
b2—  40c=o? 

EXERCISES 

1.  State  without  solving  the  equations,  whether  the  roots 
of  the  equations  1-5,  §  218,  are  real. 

2.  Determine  the  value  of  m  for  which  the  chords  in  Fig.  236 
actually  exist. 


FIG.  236 

We  have:   (w*)2+42  =  (3*-4)2  +  52.     Why? 
i.  e.,  w2#2  +  i6=9#2  —  24^+16  +  25 
(m*  —<j)x2  +  24*— 25  =o. 


Second-Year  Mathematics 


In  this  case    a  =  m2—  9 
6  =  24 
c=-25 

That  the  chords  may  actually  exist  x  must  be  a  real  num- 
ber. This  will  be  true  if 

(24)2-4(w2-9)(-25)>o 
i.e.,  576  +  ioofw2  — 9oo>o 
or  if  w2  >if£ 
i.  e.,  if  w>f  or  w<  —  §-.     Why? 

3.  The  meaning  of  1I}12,  dx,  and  d2  being  as  in  Exercise  5, 
p.  198,  determine  the  value  of  m  for  which  the  chords  have 
real  existence: 


/i 

1, 

d, 

d, 

(l).. 

2  1/  x  —  m 

6x  —  4 

e 

4 

(2).. 

8 

i2m 

* 

5* 

(3)  

6x-S 

2X+6 

2m 

i 

(4).. 

2X—8 

IO#+  2 

•j/  2x-\-  yn 

-j/  m  —  x 

*  In  this  equation  b  is  o. 

PROBLEMS  AND  EXERCISES 
220.  Solve  the  following  problems  and  exercises: 
i .  Construct  a  triangle  ABC,  the  sides  a  and  b  and  the 
angle  A,  opposite  one  of  them,  being  given  (Fig.  237). 


A' 


C 


FIG.  237 


FIG.  238 


Draw  a  line  A  C  equal  in  length  to  b  (Fig.  238).     At  the 
extremity  A  of  A  C  construct  an  angle  equal  to  the  given 


Inequalities  in  Triangles  and  Circles 


203 


angle  A .  Now,  with  C  as  a  center  and  a  as  a  radius,  strike  an 
arc  meeting  the  second  side  of  Z.A  at  two  points  B  and  B'. 
Both  of  the  triangles  A  C  B  and  A  C  B'  have  the  required  parts. 

2.  Draw  the  perpendicular  C  B"  from  C  to  A  B  (Fig.  238). 

How  many  triangles  would  have  been  formed  in  case 
a<CB"?  Why? 

How  many  if  a  =  C  B"  ?    Why  ? 


FIG.  240 

3.  Prove,  that  Z.B   in  /^A  B  C  must  be  acute. 

4.  Prove,  that  Z.B'  in  /\A  B'C  must  be  obtuse. 

5.  Prove,  that  whenever  the  construction  in  Problem   i 
furnishes  two  triangles,  one  of  those  must  be  an  acute-angle 
triangle,  and  the  other  one  an  obtuse-angle  triangle. 


2O4 


Second-  Year  Mathematics 


FIG.  241 


6.  Given  two  triangles  ABC  and  D  E  F,  having  A  B  =D  E, 
B  C=E  F  and  Z.C  opposite  A  B  =  Z.F,  opposite  D  E. 

7.  Prove  by  the  method  of 
superposition  and  by  the  use  of 
Exercises  i  to  6,  that  A  A  B  C  = 
AD  E  F,  in  case  Z_A  and  /_D 
are  both  acute,  both  obtuse,  or 
both  right  angles. 

221.  The  distance  from  a 
point  P  to  a  circle  whose  center 
is  O,  is  the  length  of  PA,  that 
portion  of  the  line  P  O  which 
lies  between  P  and  the  point 
where  P  O  meets  the  circle  (see  Fig.  239). 

Hence  the  center  O2  of  circle  C2  (Figs.  240  and  241)  is  at 
a  distance  r2  from  circle  Ct;  the  center  Ox  of  circle  Cz  is  at 
a  distance  rt  from  circle  C2. 

EXERCISES 

1.  Find  the  locus  of  the  points  whose  distance  from  a 
given  circle  C  is  equal  to  a  given  length. 

2.  Find  the  locus  of  the  center  of  the  circles  of  given  radius 
a,  which  are  externally  or  internally  tangent  to  a  given  circle 
(see  Fig.  242). 

3.  Determine  the  points  which  shall  lie  at  a  given  distance 
from  both  circle  A  and  circle  B  (Fig.  242). 

Where  are  all  the  points  collected  that  have  the  required  distance 
from  circle  A  ? 

Where  are  all  the  points  collected  that  have  the  required  distance 
from  circle  B  ? 

Where  are  then  found  the  points  which  have  the  required  distance 
from  both  circle  A  and  circle  B  ? 

4.  Construct  as  many  circles  as  possible  with  radius  of 
i  inch,  which  shall  be  tangent  at  the  same  time  to  circle  d 


Inequalities  in  Triangles  and  Circles 


205 


of  radius  5  inches  and  to  circle  C2  of  radius  4  inches,  when  the 
center  Oz  and  O2  of  the  circles  Cr  and  C2  are  6  inches  apart. 

5.  Determine  what  relation  must  hold  between  the  radii 
fj  and  r2  of  circles  Ct  and  C2  the  distance,  d,  between  their 
centers  and  a  line  p,  in  order  that  there  may  be  8  circles  of 
radius  p  tangent  to  both  Cr  and  C2. 


FIG.  242 

6.  Determine  under  what  conditions  there  will  be  6  circles; 
2  circles;  none,  of  the  nature  described  in  Exercise  5. 

7.  Can  there  ever  be  4  circles  of  that  kind? 


Exercises  for  Algebraic  Solution 
222.  Solve  the  following  quadratics: 

(1)  r2—  gr—  36=0  (6) 

(2)  /2  +  i5*=-44  (7) 

(3)  s2-72=6s  (8) 

(4)  3W2=6  —  7w  (9) 


=  28^  —  io;y2  +  5 


(5) 


=—  1#2  —  20 


(10) 


2o6  Second-Year  Mathematics 

223.  Solve  the  following  quadratics  for  y: 

(1)  i4^  =  3-8j2  (5)  8^2+8^  +  2c2  =  -i9C2-6>-2 

(2)  a-y2  =  (i—a)y  (6)  y3+py+q=o 

(3)  2i62  =  23&j— 6y2  (7)  aj2+&^+c=o 

(4)  28m2  =  -i7&j+3>'2          (8)  xy*+zy=—s. 

224.  Determine  by  inspection  the  nature  of  the  roots  of  the 
following  quadratics: 

(1)  *2+6*  +  5=o  (6)  2$p2  +  i=o 

(2)  4y2-gy  +  2=o  (7)  /2~ 

(3)  4W2—  8w+4=o  (8)  s2  — 

(4)  3y2  +  7?  +  2=o  (9)  ^ 

(5)  r2-i=o  (10)  8/2- 


CHAPTER  VI 
AREAS  OF   POLYGONS 

225.  To  measure  the  size  of  a  plane  surface  is  to  compare 
that  surface  with  a  standard  surface,  a  unit  of  measurement. 
A  square  is  used  as  the  unit  of  measurement  for  plane  sur- 
faces.   The  number  of  such  units  contained  in  a  given  sur- 
face expresses  the  area  of  the  surface  in  terms  of  that  unit. 

Thus,  if  a  rectangle  has  sides  of  4  in.  and  6  in.,  it  is  easily 
seen  that  a  square  inch  may  be  laid  off  on  it  24  times.  We 
say  that  24  expresses  the  area  of  the  rectangle,  in  terms  of 
square  inches. 

226.  The  areas  of  polygons  may  be  found  by  dividing  the 
polygons  into  triangles,  as  in  Figs.  243  and  244,  and  then 
adding  the  areas  of  the  triangles. 


FIG.  243 


FIG.  244 


The  Area  of  the  Triangle 

227.  Thus  it  is  seen  that  it  is  important  to  know  how  to 
compute  the  area  of  a  triangle.  Several  formulas  will  be  worked 
out  by  which  the  area  of  a  triangle  can  be  found.  The  parts 
of  the  triangle  that  are  known  will  enable  the  pupil  to  decide 
which  formula  can  be  used. 


207 


208  Second-Year  Mathematics 

Fig.  245  shows  two  triangles  each  one-half  of  a  parallelo- 
gram. Thus  if  we  are  able  to  find  the  area  of  a  parallelogram, 
the  area  of  the  triangle  is  found. 


FIG.  245  FIG.  246 

Fig.  246  shows  how  the  area  of  a  parallelogram  can  easily 
be  compared  with  the  area  of  a  rectangle.  The  question 
arises,  whether  the  rectangle  and  the  parallelogram  of  Fig.  246 
are  equal  to  each  other,  and,  more  generally,  under  what  con- 
ditions such  figures  are  equal  to  each  other.  This  question 
is  answered  by  Exercise  i,  §  229. 

If  we  know  how  to  compare  parallelograms,  we  can  com- 
pare parallelograms  with  rectangles  because  a  rectangle  is  a 
special  case  of  a  parallelogram. 

228.  We  have  thus  the  following  chain  of  problems  leading 
to  the  problem  of  computing  the  area  of  the  triangle: 

(1)  to  compare  parallelograms 

(2)  to  compare  a  parallelogram  and  a  rectangle 

(3)  to  compute  the  area  of  a  rectangle 

(4)  to  compute  the  area  of  a  parallelogram 

(5)  to  compare  a  parallelogram  and  a  triangle 

(6)  to  compute  the  area  of  a  triangle. 

PROPOSITION  I 

229.  Theorem:     Parallelograms   having  equal   bases  and 
equal  altitudes  are  equal. 

Hypothesis:  In  parallelograms  ABCD  and  EFGH 
(Fig.  247,  (i)),  h=h'.  AB=HG; 


Areas  of  Polygons  209 

Conclusion:    ABCD=EFGH. 

Proof:  Place  ABCDonEFGH  making  A  B  coincide 
with  H  G. 

Then  D  C  must  fall  in  line  E  F,  in  position  D'C',  for  the 
altitudes  of  the  parallelograms  are  equal. 

XL 


(2) 


Prove  AED'H^AFC'G.  If  triangle  E  D'H  is  sub- 
tracted from  the.quadrilateral  H  E  C'G,  H  D'C'G  ( =  A  B  C  D) 
remains.  If  triangle  F  C'G  is  subtracted  from  the  quadri- 
lateral H  E  C'G,  E  F  G  H  remains. 

.'.  Parallelogram  A  B  CD=parallelogram  E  F  G  H. 

Prove  the  theorem  for  Fig.  247  (2). 

Hence,  parallelograms  having  equal  bases  and  equal  alti- 
tudes are  equal.  Q.E.D. 

EXERCISES 

1.  Prove  that  the  area  of  a  parallelogram  equals  that  of  a 
rectangle  having  the  same  base  and  altitude  (see  Fig.  248). 

2.  To  find  the  number  of  surface  units  (area)  in  a  rectangle, 
•when  the  number  of  linear  units  in  the  base  and  in  the  altitude 
are  known. 


Second-  Year  Mathematics 


On   squared  paper   draw  a  rectangle   with   corners  at  the  points 

(*>  !)»  (5>  !).  (5,  3).  C1,  3)  (see  FJg-  249)- 

By  counting  the  squares  in  A  B  C  D  (Fig.  249)  find  the  area  in 
terms  of  the  squares  of  the  paper. 


FIG.  248 

3.  On  squared  paper  draw  rectangles  having  the  following 
dimensions: 


Base          

i 

7 

I? 

10 

"LA. 

Altitude  

c 

4 

•} 

2 

7 

4.  Using  as  surface  unit  a  square  whose  side  is  the  linear 
unit,  find  the  area  of  the  rectangles  in  Exercise  3. 

230.  How  can  the  number  of  surface  units  in  a  rectangle 
be  obtained  from  the  number  of  linear  units  in  the  base  and 
altitude  ? 


« 

3) 

'5,3) 

A 

2 

D 

C 

| 

I 

^/^ 

i 

4 

l     i    4^  i 

>      6 

FIG.  249 


b-. 

FIG.  250 


Let  &  denote  the  length  of  the  base  and  h  the  length  of  the 
altitude  of  a  rectangle  A  B  C  D  (Fig.  250). 

Then  the  linear  unit  can  be  laid  off  b  times  on  A  B  and 
h  times  on  A  D. 


Areas  of  Polygons 


211 


Through  the  points  of  division  of  A  B  draw  lines  parallel 
to  A  D,  as  E  E',  F  F',  etc.  (Fig.  251) 

Then  rectangle  A  B  C  D  is 
divided  into  b  congruent  rec- 
tangles R.  Through  the  points 
of  division  of  'A  D,  draw  lines 
parallel  to  A  B,  as  K  K',  L  L', 
etc 

Then  each  rectangle  R  is 
divided  into  h  congruent 
squares.  Denoting  the  num- 


*-* 

£.'     / 

m    1 

C 

1 

1 



1 



A, 

R 

/' 

1  L 

!* 

K' 

i 

A     E    F 


_'_  t 

FIG.  251 


her  of  squares  in  A  B  C  D  by  S,  it  follows  that 

S  =  b-h.  (a) 

This  equation  may  be  stated  in  words  thus: 

PROPOSITION  II 

231.  Theorem:    The  area  of  a  rectangle  equals  the  product 
of  the  base  by  the  altitude. 

EXERCISES 

i.  Verify  formula  (a)  for  6=4,  h  =  i  .4. 
Lay  off  A  6=4  cm.  and  A  D  =  i  .4  cm.  (Fig.  252). 


M»f 

i 

% 

—  . 

j 

i 

/»                                          A  -° 

4 4 » 

FIG.  252 

Then  A  B  C  D  =  A  B  F  E +E  F  C  D.     (What  axiom  ?) 
(i)  Counting  squares,  we  find 

A  B  F  £=4  sq.  cm.,  E  F  C  D=4X14(r  sq.  cm. 

ABC  D=4  sq.  cm. +4X1*0  sq.  cm. 

=4(1  +T4ff)  sq.  cm.  =  5. 6  sq.  cm. 


212 


Second-  Year  Mathematics 


(2)  By  formula  (a)  we  find  S=b  •  ^=4X1 .4  =  5.6. 

2.  Verify  formula  (a)  for  6=3.8,  /z  =  5;  for  6=6,  h  =  2.6; 
for  6=3,  A =4. 2. 

3.  Verify  formula  (a)  for  6=4.6,  ^  =  2.4  (see  Fig.  253). 


FIG.  253 


.  cm. 
=  (8  +  1.2+1.6+.  24)  sq.  cm.,  etc. 

4.  Verify  formula  (a)  for  6  =  1  .4,  ^=0.32. 

232.  Exercises  i,  2,  3,  and  4,  §  229,  show  that  formula  (a) 
holds  when  6  and  h  are  integers  (whole  numbers).. 

Exercises  i,  2,  3,  and  4,   §  231,  show  that  formula  (a) 
holds  also  when  6  and  &  are  decimal  fractions. 

233.  Now  let  6  and  h  be  irrational  numbers,  such  as 

6  =  1/12=3.464101  .  .  .  .  , 
and  ^  =  1/27  =  5.196152  ..... 

If  the  theorem  holds  in  this  case,  we  should  find: 


lr2Xl27  =  l22  •  3  •  33=i8.     (?) 

Taking  for  6  and  h  the  values  given  in  I,  II,  III,  and  IV 
below,  we  find  the  corresponding  values  of  5  by  multiplying. 


Areas  of  Polygons 


213 


6 

h 

5-6XA 

I  

•?  .464 

<.  .  106 

17.008044 

II  

3  .4641 

?  .1061 

17.00081001 

Ill  

3  .46410 

<;  .  10611; 

1  7.  00008  32  1  <; 

rv  

•?  .464101 

s  .  1061^2 

17  QQQQQS^Q?1^ 

We  see  that  by  taking: 

I,  b  and  h  accurate  to  within  —  —  ,  5  differs  from  18  by  less 

I,OOO7  » 


,  -S1  differs  from  18  by 


S   differs  from   18 


differs  from  18 


II,  b  and  h  accurate  to  within 

less  than  i 

III,  &  and  /&  accurate  to  within 

1,0 

by  less  than  —  —  . 
10,000 

IV,  b  and  h  accurate  to  within  —  ^  —  , 

I  ,OOO,OOO 

by  less  than  —  -  —  . 
t  100,000 

By  taking  b  and  h  to  a  sufficiently  large  number  of  decimal 
places,  the  difference  between  S  and  18  may  be  made  as  small 
as  may  be  wished. 

To  prove  from  these  facts  that  our  theorem  holds  true 
when  b  and  h  are  irrational  numbers  would  require  a  fuller 
discussion  of  such  numbers  than  is  desirable  here.  Sufficient 
has  been  shown,  however,  to  make  us  feel  that  this  conclusion 
is  correct,  for  the  inexactness  of  the  result  seems  to  come  from 
our  inability  to  express  exactly,  without  the  root  sign,  the 
dimensions  of  the  rectangle,  and  not  because  the  theorem  is 
untrue. 

i.  Make  a  similar  discussion  for  6  =  1/5  an(^  ^  =  1/7;  f°r 
&  =  j/2  and  fc  =  i/3. 

EXERCISES 

i.  Prove  that  any  two  parallelograms  are  to  each  other  as 
the  product  of  their  bases  and  altitudes. 


214  Second-Year  Mathematics 

2.  Prove  that  the  areas  of  parallelograms  having  equal 
bases  are  to  each  other  as  their  altitudes. 

3.  Prove  that  the  areas  of  parallelograms  having  equal 
altitudes,  are  to  each  other  as  their  bases. 

4.  Prove  that  the  area  of  a  parallelogram  is  equal  to  the 
product  of  the  base  and  altitude.     (Use  Exercise  2.) 

5.  Prove  that  the  area  of  a  triangle  is  one-half  the  product 
of  the  base  and  altitude.     (Use  Exercise  4  and  Fig.  254.) 

Thus  the  area  of  a  triangle  can  be  computed  if  the  base  and  the 
altitude  are  known. 


6.  Show  that  the  area  of  a  triangle  can  be  expressed  in 
terms  of  any  two  sides,  and  the  sine  of  the  included  angle. 

In  Fig.  254  let  a  =  side  D  B,  6  =  side  D  A;    then    (§  137,  p.   104) 
show  that  h  =  a  sin  D.     .'.  Area  A  B  D  =  %b  •  h  =  %b  •  a  sin  D. 

7.  Prove  that  triangles  having  equal  bases  and  equal  alti- 
tudes are  equal. 

8.  Prove  that  the  areas  of  triangles  are  to  each  other  as  the 
products  of  the  bases  and  altitudes. 

9.  Prove  that  the  areas  of  triangles  having  equal  bases  are 
to  each  other  as  the  altitudes. 

10.  Prove  that  the  areas  of  triangles  having  equal  altitudes 
are  to  each  other  as  the  bases. 

u.  To  bisect  a  triangle  by  a  line  drawn  from  any  vertex 
to  the  opposite  side. 
See  Exercise  7. 


Areas  of  Polygons  215 

12.  To  divide  a  triangle  into  three  equal  triangles  (trisect) 
by  lines  from  any  vertex  to  the  opposite  side. 

234.  The  area  of  a  triangle  can  be  expressed  in  terms  of 
the  perimeter  of  the  triangle  and  the  radius  of  the  inscribed 
circle,  by  dividing  it  into  three  triangles  whose  common  alti- 
tude is  the  radius,  and  whose  bases  are  the  sides. 

i.  Prove  that  the  area  of  a  triangle  is  equal  to  one-half  the 
perimeter  times  the  radius  of  the  inscribed  circle  (see  Fig.  255). 

c 


FIG.  255 

The  area  of  a  triangle  may  i>e  expressed  in  terms  of  the  sides  of  the 
triangle  and  the  radius  of  the  circumscribed  circle. 

2.  Prove  that  the  area  of  a  triangle  is  equal  to  the  product 
of  the  three  sides  divided  by  four  times  the  radius  of  the  cir- 
cumscribed circle.  (See  §  189.) 

The  area  of  a  triangle  can  be  expressed  in  terms  of  the  sides 
alone,  but  in  the  proof  of  the  formula,  the  following  theorem 
is  needed: 

PROPOSITION  III 

235.  Theorem:  The  square  on  the  hypotenuse  of  a  right 
triangle  is  equal  to  the  sum  of  the  squares  on  the  sides  including 
the  right  angle. 

Hypothesis:  ABC  (Fig.  256)  is  a  triangle  having  a  right 
angle  C;  SI}  S3,  and  5"  are  the  squares  on  the  sides  a,  b,  and 
c,  respectively; 


2l6 


Second-Year  Mathematics 


Conclusion:    S=SI+S2. 

Proof:  Draw  CD_l_AB,  dividing  5  into  rectangles  RT 
and  R2.  Draw  AE  and  C  F.  Show  that  triangle  EBA 
and  Sj  have  equal  bases  and  altitudes. 


r   D 


FIG.  256 

Then  triangle  E  B  A  =  J5,. 
Similarly,  prove  that  triangle  FB  C  =  J 
But  ABE^FB  C. 
For     EB=BC     (?) 
AB=BF     (?) 
ZABE=ZFBC     (?) 
From  (i),  (2),  (3)  it  follows  that  ^  =  1 


Similarly,  draw  B  G  and  C  H,  and  prove  Sa=R3 
Therefore  S=St+Sa.     Q.E.D. 


(i) 
(2) 
(3) 


(4) 


Areas  of  Polygons  217 

This  theorem  is  called  the  "theorem  of  Pythagoras"  after  the 
Greek  mathematirian  Pythagoras,  569-500  B.  c.,  who  first  proved  it. 
It  is  one  of  the  most  famous  propositions  of  geometry,  and  numerous 
proofs  are  on  record.  The  proof  which  precedes  is  due  to  Euclid.  A 
few  other  proofs  are  given  later  (see  §  238). 

EXERCISES 

1.  The  area  of  a  rectangle  is  i6y3  —  4y,  and  the  altitude  is 
4jy.     The  diagonal  is  10,  find  y,  and  the  sides  of  the  rectangle. 

2.  The  diagonal  of  a  rectangle  is  20  and  one  side  is  16. 
If  x3  +4*  denotes  the  area,  what  does  x  equal  ? 

3.  The  area  of  a  rectangle  is  300  and  one  side  is  15.     Find 
the  perimeter. 

4.  The  perimeter  of  a  rectangle  is  84,  and  the  sides  are  to 
each  other  as  3  to  4.     Find  the  diagonal. 

5.  The  area  of  a  rectangle  is  48,  and  the  base  is  8.     Find 
the  diagonal. 

6.  The  altitude  of  a  rectangle  is  24  and  the  area  is  768. 
Find  the  diagonal. 

7.  The  area  of  a  rectangle  is  2C3  —  2C,  and  the  base  is  20. 
Find  the  diagonal. 

8.  The  diagonal  of  a  rectangle  is  15  and  the  altitude  is  9. 
Find  the  area. 

9.  One  side  of  a  rectangle  is  2X.    The  diagonal  is  x2  + 1 ,  and 
the  area  is  12.     Find  x,  and  the  sides,  and  the  diagonal. 

Use  the  factor  theorem. 

PROPOSITION  IV 

236.  Problem:  To  find  the  area  of  a  triangle,  in  terms  of 
its  sides,  to  be  1/5(5— a) (s—b) (s— c). 

Hypothesis:  In  triangle  ABC  (Fig.  257)  the  sides  a,  b, 
and  c  are  known. 


2l8 


Second-  Year  Mathematics 


To  express  the  area  in  terms  of  a,  b,  and  c. 
Proof:  Area  A  B  C  =  J6  •  h  (?). 


(i) 


C 


This  gives  the  area  in 
terms  of  one  side  and  the 
altitude  h,  which  is  not 
known.  Therefore  we  will 
express  h  in  terms  of  the 
sides,  to  substitute  for  h 


< 0 »"      in  equation  (i). 

FIG.  257 

h2=c2-(b-a')2  (why?).  (2) 

h*=a2-a'2  (why?).  (3) 

We  must  now  eliminate  a',  which  is  not  one  of  the  three  sides. 
Subtracting,  c2-a2+a'2-(b-a')2=o.  (4) 

Therefore  c2—  a2—  b2  +  2ba'=o.  (5) 

b2-c2+a2 


Solving  for  a',  we  find  a'=- 


(6) 


Substituting  in  (3)  the  value  of  a'  found  in  (6),  we  get 

(7) 


Equation  (7)  expresses  A2  in  terms  of  the  sides  a,  b,  and  c. 

We  could  now  substitute  the  value  of  h  in  equation  (i) 
and  have  a  formula  for  the  area  of  A  B  C  in  terms  of  a,  b, 
and  c.  But  in  order  to  get  a  more  symmetrical  result,  the  value 
of  h  in  (7)  will  be  changed  in  form  before  substituting  in  (i). 
b2-c2+a2\/  b2-c2  + 


From  (7)  h2  = 


\            2b 

;r    a*  /' 

+a2     2ab—  b2+c2  —  a2 

2b 

(a+b)2-c2 

2b 
c2-(a-b)2 

2b 

2b 

,   . 
,  (why?) 


2b 


(c+a-b)(c-a+b) 

2b 


_ 
Why? 


Areas  of  Polygons  219 

Denote  (a  +  6+c)  by  2s. 

Then  a+b—c  =  2s  —  20  =  2(5—  c)  \ 

—  a  =  25  —  2a  =  2(i-—  a)  '  (9) 

—b  =  2s  —  2b  =  2(s—b')  / 
Substituting  (9)  in  (8) 

2(s—c)-  2s  •  2(5—6)  •  2(3—  a)_45  •  (5—  a)(s—  b)(s  —  c) 
~  ~~ 


Therefore  h=jVs(s -a) (s-b) (s -c).     Why?  (10) 

Substituting  (10)  in  (i), 


ABC  =  $6  •  \ 
b 


Therefore  A  B  C  =  \s(s-a)(s-b)(s-c).      (i i )      Q.E.D. 

Since  a  +  b  +  c  =  2s,  then  s  =  %  (a  +  b  +  c). 

Formula  (n)  expresses  the  square  root  of  the  product  of  the  half- 
sum  of  the  three  sides,  and  the  half-sum  diminished  by  each  side  in 
succession. 

Problems 

1.  The  sides  of  a  triangle  are  3,  5,  and  6.     Find  the  area. 
Using  the  formula  (n)  of  §  236, 

the  area  =  1/7  •  (7—3)  (7—5)  (7—  6)  =  1/7  -4  •  2  •  i  =21/14,  or  7.482 
approximately. 

2.  The  sides  of  a  triangle  are  34,  20,  and  18.     Find  the 
area. 

3.  The  sides  of  a  triangle  are  10,  6,  and  8.     Find  the  area. 

4.  The  sides  of  a  triangle  are  90,  80,  and  26.     Find  the  area. 

5.  The  sides  of  a  triangle  are  70,  58,  andi6.     Find  the  area. 

AREAS  AND  ALTITUDES 

237.  The  altitudes  of  the  triangle  A  B  C  to  the  sides  a,  b, 
and  c  are  ha,  hf,,  and  hc,  respectively  (Fig.  258). 


Second-  Year  Mathematics 


=^V  s(s—a}(s—b}(s—c)    (i)    (See  §  236,  Formula  (10).) 


ha=-Vs(s-a)(s-b)(s-c)     (2) 
a 


hc=-Vs(s-a)(s-b}(s-c).     (3) 

G 


How  can  (2)  and  (3)  be  obtained  from  (i)  by  inspection? 


Problems 

i.  In  the  triangle  ABC,  a  =  io,  6  =  17,  c  =  2i.     Find  ha. 

2 


2     /  - 

ha=-v  s(s—  a)(s  —  b)(s  —  c) 


1)  =24 

s—  a  =  i4,  s—b  —  j,  s—c=3. 
Substitute  these  values  in  the  formula,  and 


24  •  14  •  7  '  3=4  -3-2.  2-  7-  7-  3  = 
-9.4-  49=i(2  •  3  '  2  •  7)=¥  =  i6f 

2.  Find  the  area  and  the  altitudes  of  each  of  the  following 
triangles: 

(0  0  =  35.  6  =  29,  c=  8        (4)  a  =  i5,  6  =  20,  ^  =  25 

(2)  0  =  70,  6=65,  c=  9        (5)  a=  8,  6=  8,  c=  8 

(3)  a=45>  ^=40,  ^  =  13        (6)  a  =  i7,  6  =  10,  c=  9 


Areas  of  Polygons  221 

3.  The  sides  of  a  quadrilateral  are  as  follows:    A  6=29, 
B  C=8,  C  D=-28,  D  A  =  2i,  and  the  diagonal  A  C=3o.     Find 
its  area,  and  the  distance  from  D  to  A  C. 

4.  The  area  of  a  triangle  in  terms  of  its  sides  and  the 
radius  of  the  circumscribed  circle  is  —  -  (see  §  234,  2).     Let 

T  denote  the  area  of  the  triangle,  then  T—  —  .     Solving  this 

equation  for  r,  we  get  r=—=..     Using   this   formula    for    r, 

4* 

find  the  radius  of  the  circles  circumscribed   about  the   tri- 
angles of  Problem  2. 

5.  To  find  the  altitude  and  area  of  an  equilateral  triangle  in 
terms  of  its  side. 

Let  each  side  be  a.    The  altitude  h,  bisects  the  base.    Why  ? 
&2=a2-($a)2=a2-£a2  =  £a2          The  area  S=$a*h 

-*  •  VI 


5=io 

A  =-1/3  or  iai/3".  S=laai/3- 

6.  Find  the   areas  of  the  following  equilateral  triangles 
(a=side):  (i)  a  =  i2,  (2)  a  =  io,  (3)  a=4,  (4)  a=8,  (5)  a=c+d, 
(6)  a  =  2mn. 

Compare  your  answer  of  (4)  with  2  (5). 

7.  Find  the  altitudes  of  the  triangles  of  Problem  6. 

8.  Find  the  side  of  an  equilateral  triangle  whose  area  is 
(i)  ^1/3        (2)  25!/3        (3)  i2        (4)  101/3. 

9.  Find  the  altitude  of  each  of  the  triangles  in  Problem  8. 

10.  Find  the  area  of  a  regular  hexagon  whose  side  is 
(!)  12        (2)  8        (3)  10        (4)  7        (5)   -6. 

See  FYM,  p.  360,  6. 


222 


Second-Year  Mathematics 


238.  Prove  the  Pythagorean  proposition  by  each  of  the 
following  figures  (see  footnote,  p.  216).  (In  each  figure  C  is 
the  vertex  of  the  right  angle  of  the  right  triangle  ABC.) 

Figs.  259  and  260.  From  the  dimensions  given  in  the 
figures,  find  the  areas  of  the  triangles,  and  the  square  on  c 
by  subtraction  or  addition.  Use  the  algebraic  method. 

b   K  £  a    f 


FIG.  260 


FIG.  261 


Fig.  261.     G  C  is  a  square  on  a+&.     GE- 
A  E,  ED,  B  D.     Prove  A  B  D  E  a  square, 
two    rectangles    G  O 


=D  F=a  Draw 
Show  that  the 
and  O  C  are 

equivalent  to  the  four  right  triangles. 
Subtract  each  in  turn  from  G  C. 

Fig.  262.  Complete  the  rectangle 
CJKT.  DrawCK.  DrawCG||BE. 
Show  that  K  C  G  is  a  straight  line. 
Prolong  D  A  to  HJ,  and  EB  to  KT. 
Compare  A I  K  C  with  the  square 
AHJC,  and  KB 
with  C  S  and  F  E. 
Fig.  263.  Con- 
struct A  B  D  E  a  square.  Through  E 
draw  aline  ||  to  AC.  Draw  JLS  as  indi- 
cated. Show  that  BC=HG,  and  that 
D  G  and  C  K  are  squares,  and  that  the  pIG 

right  triangles  are  equal. 

Fig.  264.    Construct  E  H  ||  B  C,  DK  ||  BF.    Draw  C  K  L 
and  prove  it  to  be  ||  to  B  E.     Compare  A  C  K  D  with  A  L,  etc. 


Areas  of  Polygons 


223 


Fig.  265.  Construct  DEF=ABC.  Draw  CG  and  CJ. 
Prove  G  C  J  a  straight  line,  and  the  quadrilaterals  G  H  I  J, 
GABJ,  CADF,  CBEF  equal,  etc. 


FIG.  264 

Fig.  266.  Find  b  in  terms  of  a  and  c  applying  the  law  of 
the  secant  and  tangent.     Use  the  algebraic  method. 

239.  It  has  now  been  proved  in 
many  ways  that  the  square  on  the 
hypotenuse  of  a  right  triangle  equals 
the  sum  of  the  squares  on  the  sides 
including  the  right  angle. 


FIG.  266 


FIG.  267 


Imagine  the  angle  ABC 
(Fig.  267)  to  increase,  leaving 
the  lengths  of  the  sides  A  B 
and  B  C  unchanged.  Then 
the  squares  on  A  B  and  B  C 
are  not  changed  in  size,  but 
as  the  end-points  A  and  C 
of  A  B  and  B  C,  are  farther 
apart,  the  square  on  AC 
increases. 

Therefore  in  an  obtuse- 
angled  triangle,  as  Fig.  268, 


224 


Second-Year  Matliematics 


the  square  on  the  side  opposite  the  obtuse  angle  is  greater 
than  the  sum  of  the  squares  on  the  other  two  sides. 


FIG.  268 


FIG.  269 


240.  In  a  similar  way,  by  decreasing  angle  ABC,  until 
it  becomes  acute,  as  in  Fig.  269,  we  find  that  the  square  on 
the  side  opposite  the  acute  angle  is  less  than  the  sum  of  the 
squares  on  the  other  two  sides. 

EXERCISES 
i.  What  is  the  relation  between  a2,  b2,  and  c2  (Fig.  270) 

whenZ£<9o°? 
when  Z-B>9o°? 

2.  How  does  the   Z-6  com- 
pare with  a  right  angle  when — 


c 

FIG.  270 


(2)  b2—  a2+c2  ? 

(3)  62<a2+c2? 

3.  Is  the  Z.B  >  or  <  90°  when  a,  &,  and  c  are,  respectively, 


Areas  of  Polygons  225 

(1)  3>  5^4  (5)  9.  14,  12 

(2)  5,  13,  12  (6)  z2-y2,  z2+y2,  2zy 

(3)  5,8,6  (7)  x2-i,x2,  2X 

(4)  6,  12,  8  (8)  2k*-3h2,  2(k2+h2),  2^hk? 

241.  The  following  two  theorems  will  show  by  how  much 
the  square  on  one  side  of  a  triangle  differs  from  the  sum  of 
the  squares  on  the  other  two  sides. 

Problem :  Let  /_  B  be  an  acute  angle  of  triangle  ABC 
(Fig.  271). 

c         c 


Let  C  D  be  perpendicular  to  A  B. 
To  prove:    b2=a2+c2  —  2caf. 
Proof:   b2=h2  +  (c-a')2     (why?) 

a2=h2+a'2. 

Subtracting  b2-a2  =  (c-a')2-a'2=c2-2ca'+a'2-af2. 
Therefore  b2—a2=c2  —  2ca'. 
Solving  for  b2,  b2  =a2  +c2  -  2ca'.     Q.E.D. 
This  shows  that  the  product  2ca'  is  the  amount  by  which 
aa+c*  exceeds  b2. 

a'  (Fig.  271)  is  called  the  projection  of  a  upon  c. 

242.  The  projection  of  a  given  line-segment  upon  a  straight 
line,  is  the  segment  of  the  line  included  between  the  perpen- 
diculars to  the  line  which  pass  through  the  extremities  of  the 
given  line-segment. 


226 


Second-  Year  Mathematics 


In  Fig.  272  A'B'  (or  A  B')  is  the  projection  of  A  B. 

£  £ 


/!' 


C 


A 

FIG.  272 


The  problem  of  §  241  may  now  be  stated  thus: 

PROPOSITION  V 

243.  Theorem:  In  a  triangle  the  square  on  the  side  oppo- 
site an  acute  angle  is  equal  to  the  sum  of  the  squares  of  the 
other  two  sides,  diminished  by  two  limes  the  product  of  one  of 
these  two  sides  and  the  projection  of  the  other  upon  it. 

EXERCISES 

i.  b  is  opposite  an  acute  angle,  and  c'  is  the  projection  of 
c  on  a  (Fig.  273).  Test  Proposition  V  with  the  figures  below, 
i.  e.,  show  that  b2=a2+c2  —  iac'. 


h, 


/oN    r, 
'       <Z  ^.*- 


C  C 

FIG.  273 

2.  a,  b,  and  c  are,  respectively,  5,  10,  and  9.     (i)  Is 

90°  ?*  (2)  Find  a'  the  projection  of  a  on  c,  and  c'  the  pro- 
jection of  c  on  a.  (3)  a2+c2  exceeds  b2  by  how  much? 
(4)  Compare  2a'c  and  2acr. 

3.  a,  b,  and  c  are,  respectively,  5,  13,  12.     (i)  Find  a'  andc' 
as  in  Exercise  2.     (2)  Is  Z^trgo0?     (3)  Does  Proposition 
V  hold  true  when  Z_B  is  a  right  angle  ? 

*  ^  means  less  than  or  greater  than. 


Areas  of  Polygons  227 

4.  Find  a'  and  c',  when  a,  b,  and  c  are  respectively — 

(1)  7,  9,  6  (3)  *2-;y,  *a~3y,  2*i/y 

(2)  II,  15,  12  (4)    Z2-I,  22-3,   2Z. 
Check  each  in  the  equation  2a'c  =  2ac'. 

5.  Show  from  Fig.  273  that  c'=c  cos  B  (§  137,  p.  104),  and 
that  the  formula  of  Exercise  i  may  be  written 

b2=a2+c2-2accos  B* 

6.  Similarly  write  the  equation  for  a2;  for  b2. 

PROPOSITION  VI 

244.  Theorem:  In  an  obtuse-angled  triangle  the  square  on 
the  side  opposite  the  obtuse  angle  is  equal  to  the  sum  of  the 
squares  on  the  other  two  sides,  increased  by  two  times  the  prod- 
uct of  one  of  them  and  the  projection  of  the  other  upon  it. 

Hypothesis:    Let  ZA  B  C  be  obtuse  (Fig.  274); 

C 


Proof:   b2=h3  +  (c+a')2  and  a2=h*+a'2.     Why? 
Therefore    b2-a2=c2  +  2ca'+a'2-a'2. 

b*=a2+c2 -\-2ca'.     Q.E.D. 

i .  Show  that  the  formula  of  Proposition  VI  may  be  written 
b3=a'+c'-\-2ac  cos  B. 

*  The  last  term  of  this  formula  may  be  read :   two  times  the  product  o) 
the  two  sides  into  the  cosine  o)  the  included  angle. 


228 


Second-Year  Mathematics 


PROPOSITION  VII 

245.  Theorem:  In  a  triangle  the  sum  of  the  squares  of  two 
sides  is  equal  to  twice  the  square  of  one-half  of  the  third  side 
increased  by  twice  the  square  of  the  median  to  the  third  side. 

Hypothesis:  Let  A  B  C  be  any  triangle.  In  Fig.  275  let 
m  be  the  length  of  the  median  from  C  to  A  B ; 


FIG.  275 


Conclusion:    a2+&2  =  2(c/2 

Proof:   By  Proposition  VI,  a2  = 

By  the  same,  b2  =  (c/2)3+m2  —  2(c/2)m'. 

Addkig,  a2+62=2(c/2)2  +  2w2.     Q.E.D. 


EXERCISES 

1.  The  side  opposite  an  obtuse  angle  is  b,  and  c'  is  the 
projection  of  c  upon  a  (Fig.  276).     Test  Proposition  VI  from 
the  figure,  i.  e.,  show  that  b2=a2+c2  +  2acf. 

2.  (i)  The   number   b2   exceeds   a2+c2   by   how   much? 

(2)  Compare  2a'c  and  2acf  (Fig.  276). 
3.  Find  a'  and  c'  when  a,  b,  and  c 
are,  respectively: 

(1)  5,  15,12 

(2)  6,12,8 

(3)  7,ii,8 


2S. 


Check  each  in  the  equation  2a'c  =  2ac'. 


Areas  of  Polygons 


229 


4.  Find  w^  (the  median  to  the  side  b)  when  a,  b,  and  c 
are,  respectively: 

(i)  6,  10,  8  (2)  5,  13,  12          (3)  9,  15,  12. 

5.  Find  ma  under  the  conditions  expressed  in  4. 

6.  Show  that  wc  =  iV/2(a2+62)-c2  in  any  triangle  ABC. 

PROPOSITION  VIII 

246.  Theorem:  77te  areas  of  similar  triangles  are  to  each 
other  as  the  squares  of  the  homologous  sides. 

Denoting  the  areas  of  the  given  triangles  by  T  and  T' 
(Fig.  277),  we  have 

ri  7  7  7_  7 

\b  •  h     b      h 

Y'=W^'=b'  '  h' 
i-i       (Why?)     See  §128. 


Therefore 


T'    b'    b'    b' 


Q.E.D. 


FIG.  277 
Areas  of  Polygons 

PROPOSITION  IX 

247.  Theorem:  The  areas  of  similar  polygons  are  to  each 
other  as  the  squares  of  the  homologous  sides. 

Hypothesis:  Polygon  ABC....  ^polygon  A'B'C'.  .  .  . 
(Fig.  278).  Let  P  denote  the  area  of  A  B  C  ....  and  P' 
denote  the  area  of  A'B'C'  .  .  .  .  ; 

P     d3 
Conclusion:       =-. 


230 


Second-  Year  Mathematics 


Proof:  Divide  A  B  C  and  A'B'C'  into  triangles  I,  II,  III, 
and  I',  II',  III',  respectively,  by  drawing  diagonals  from 
homologous  vertices  as  B  and  B'.  Then  I^F,  11^  IF,  etc. 
(Why?) 

C' 


FIG.  278 
Therefore:  £=£,     jp=J^, etc.     (Why?) 

But  f^W2 etc-    (why?) 

Therefore   ^=^=^  ....  etc.     (Why?) 


Therefore 


P     d 
Therefore  -= 


I+II+IH+  ....  ^11 
F+IF+IIF+  .  .  .~~IF 

Q.E.D. 


'd'2' 


(Why  ?) 


EXERCISES 

1.  The  side  of  .a  triangle  is  10  in.     Find  the  corresponding 
side  of  a  similar  triangle  having  twice  the  area. 

2.  Two  similar  triangles  have  two  homologous  sides  5  and 
15,  respectively.     What  is  the  ratio  of  their  areas? 

3.  Two  homologous  sides  of  two  similar  polygons  are  5 
and  8.     The  area  of  the  first  is  150.     Find  the  area  of  the 
second. 


231 


4.  If  one  square  is  nine  times  as  large  as  another,  what  is 
the  relative  length  of  their  homologous  sides  ? 

5.  The  area  of  a  polygon  is  6|  times  the  area  of  a  similar 
polygon.     A  side  of  the  smaller  is  4  ft.     Find  the  length  of 
the  homologous  side  of  the  larger. 

6.  Construct  a  pentagon  similar  to  a  given  pentagon  and 
equivalent  to  J  the  area. 

7.  Construct  a  square  equivalent  to  ^  of  a  given  square. 

8.  The  homologous  sides  of  two  similar  hexagons  are  9  in. 
and  12  in.,  respectively.     Find  the  homologous  side  of  a  similar 
hexagon  equivalent  to  their  sum;  equivalent  to  their  difference. 

PROPOSITION  X 

248.  Theorem:  The  areas  of  two  triangles  that  have  an 
angle  in  one  equal  to  an  angle  in  the  other,  are  in  the  same 
ratio  as  the  product  of  the  sides  including  the  equal  angles. 

Given:    AA  B  C  and  AA'B'C',  Z4  =  Z4'  (Fig.  279). 


C 

FIG.  279 


. 
o  prove: 


ACB  =  AB   •  AC 
A/B/C/-A/B/  .  A,c,  • 


232  Second-Year  Mathematics 

Proof:    Lay  off  on  A  B,  A  B"=A'B',  and  on  A  C,  A  C"  = 
A'C'.     Draw  B"C,  and  B"C". 

.'.  A  A  B"C"^  A  A'B'C'.    (  ?) 
ABC     A  B 
AIJ"C==AB"     '*     (See  *  233'  Exercise  I0'  P-  2IV 

,  AB"C"    AC" 

and  AB^C=XC    (?) 

ABC        AB-AC 


By  the  division  axiom 


B"  .  A  C"  ' 


But   A  B"=A'B',    and    A  C"=A'C' 
and  AA  B"C"^  A  A'B'C'. 

.    ABC^AB.AC  _ 

'  '  A'B'C'    A'B'  -  A'C'  ' 

Second  method,  using  §  233,  Exercise  6,  p.  214: 
ABC  =JAB  •  AC  •  sin  ,4     (?) 
A'B'C'  =  |A'B'  .  A'C'  •  sin  A'     (?) 

Since  sin  ^4=sin  A'  (see  p.  103,  §  136,  i) 
ABC^AB   »AC 
A'B'C'    A'B'  .  A7C'  ' 

EXERCISES 

1.  Two  triangles  have  an  angle  in  each  equal.     The  in- 
cluding sides  of  one  are  48  and  75,  those  of  the  other  triangle 
are  45  and  70.     Find  the  relative  areas  of  the  triangles. 

2.  Two  sides  of  a  triangular  building  are  150  ft.  and  130  ft. 
What  part  of  the  whole  building  is  included  by  50  ft.  on  the 
first  side  and  30  ft.  on  the  second  ? 

3.  A  triangular  lot  extends  60  ft.  and  80  ft.  on  two  sides 
from  a  corner.     If  a  building  is  to  front  50  ft.  on  the  first  side, 
how  many  feet  on  the  second  side  should  it  occupy  to  cover 

of  the  lot  ? 


Areas  of  Polygons 


233 


PROBLEMS 

i .  There  are  two  similar  polygons  whose  homologous  sides 
are  6  and  8.  Find  the  side  of  a  third  similar  polygon  equiva- 
lent to  their  sum. 

Construct  a  right  triangle  having  a  =  6  and  &  =  8.     Then  c  =  io.   (?) 
Construct  the  similar  polygons  Pr,  P2,  and  P3  (Fig.  280). 

2 


P,       10*' 


i2  +  82     100 
io2        100 


P.  +  Ps 

P3 

Hence  PI+P2=P3.     (?) 
Therefore  10  is  the  required  side. 


FIG.  280 

2.  There  are  two  similar  polygons  whose  homologous  sides 
are  5  and  12.     Find  the  side  of  a  third  similar  polygon  that 
is  equivalent  to  their  sum. 

3.  There  are  two  similar  polygons  whose  homologous  sides 
are  21   and  35.     Find  the  side  of  a  third  similar  polygon 
equivalent  to  their  difference. 

Construct  a  right  triangle  having  0  =  21,  and  £  =  35. 

4.  To  construct  a  figure  similar  to  two  given  similar  figures 
and  equivalent  to  their  sum  or  their  difference. 


234 


Second-Year  Mathematics 


Proof  by  the  Method  of  Analysis 

In  Fig.  281  the  four  sets  of  figures  ST,  S2,  and  S3,  taken  in 
each  set,  are  constructed  similar  figures,  of  which  a,  b,  and  c 


(I) 


a 

(2) 

b 

S3 

5, 

4 

are  homologous  sides,  equal  to  the  sides  and  the  hypotenuse 
of  a  right  triangle,  respectively;  see  Fig.  281   (5). 


Areas  of  Polygons 


235 


Then,    (i)    |i=£     (?) 
o3     c 

W  J-£    (?)  . 

(3)  ~~J5~~      ~^7~~  =  I     (?) 

/     \  C1       i     O  C*  /  ")\ 

(4)  .  .  6I+o2=o3     (?) 

(5)  and  S3—  SI=S.2.     (?) 

Hence  Fig.  281  (5)  suggests  the  method  of  construction  to 
find  the  required  homologous  side  for  the  required  figure. 

The  figure  can  then  be  constructed  similar  to  the  given 
figure. 

The  second  part  (2)  of  Problem  4  is  a  special  case  of  the  problem. 
Its  proof  follows  directly  from  the  Pythagorean  proposition,  that  is:  the 
square  on  the  hypotenuse  of  a  right  triangle  is  equivalent  to  the  sum  of 
the  squares  on  the  other  two  sides. 


JfL. 


5.  To  construct  a  figure  equivalent  to  (i)  the  sum  of  any 
number  of  given  similar  figures,  (2)  to  any  number  of  times 
a  given  figure. 

The  special  case  of  the  squares  is  given  here  because  it  is 
most  easily  comprehended. 

Given:   x,  y,  z,  and  w,  the  sides  of  given  squares  (Fig.  282). 

Required:  to  find  the  side  of  a  square  equivalent  to  the 
sum  of  the  given  squares. 


236  Second-Year  Mathematics 

(1)  Show  that  C2=x2+y2+z2+w2. 

(2)  What  relation  between  x,  y,  z,  and  w  would  make 

6.  To  find  a  line  equal  to  the  square  root  of  any  number, 
2,  3>  4,  5 

Let  x,  y,  z,  w,  .  .  .  .  in  Figure  282  each  =  i,  then  a, 
b,  c  .  .  .  .  are  equal,  respectively,  to  the  1/2,  1/3,  1/4, 
1/5 See  Fig.  282  (2). 

Is  it  possible  to  solve  Problems  i  and  2  when  the  given 
figures  are  not  similar?  For  example,  to  construct  a  figure 
equivalent  to  the  sum  of  a  square  and  a  pentagon.  Obviously 
if  the  pentagon  can  be  transformed  into  an  equivalent  square 
the  problem  can  be  solved. 

7.  To  transform  a  pentagon  into  (i)  an  equivalent  triangle, 
(2)  an  equivalent  square. 


-A  J<—  ->*r<5 * HE- 


FIG.  283 

Draw  A  F  ||  E  B  and  D  G  ||  E  C  (Fig.  283). 

Then  AEFB  =  AEAB,  AEGC  =  AEDC.     (?) 

Hence  AE  F  G=pentagon  ABODE.     (?) 


J  R  L  is  a  semicircle  and  R  K  _L  J  L.     The  area  of  Q 
1(FG)(HE).     (?) 

The  square  Q  =  the  pentagon  A  B  C  D  E.     (?) 


Areas  of  Polygons  237 

From  the  above  it  will  be  seen  that  a  polygon  of  n  sides  can 
be  transformed  into  an  equivalent  square.  Hence,  Problem  i 
is  possible  also  when  the  given  figures  are  not  similar.  In  this 
case,  however,  the  required  polygon  can  be  drawn  similar  to 
only  one  of  the  given  polygons. 

Quadratic  Equations  in  Two  Unknowns 

249.  In  the  right  triangle  ABC  (Fig.  284)  with  sides  3 
and  4,  to  construct  a  line  through  C  so  that  the  perimeters  of 
the  two  new  triangles  formed  may  be 
equal. 

Analysis:  Consider  the  problem 
solved  and  let  C  D  be  the  required 
line  through  C.  The  position  of  D 
evidently  is  determined  by  determin- 
ing A  D. 

Solution  :  Denoting  the  length  of  A  D  by  x,  and  the  length 
of  DBby;y,  3+*  +  C  D=4+;y+C  D  (?) 

x—y  =  i.  (i) 


2$.  (2) 

The  values  of  x  and  y  are  the  solutions  of  the  system  of 
equations  (i)  and  (2). 

Solving  (i)  for  x  and  substituting  in  (2), 

(i+;y)2  +  2(i+y)y+y*  =  25  (3) 

y*+y—  6=0 
(^1  =  2,        (y,=  -s 

I   #i=3  (   X2=-2 

x=—2,  ;y=—  3  satisfy  equations  (i)  and  (2)  but  do  not 
satisfy  the  conditions  of  the  problem.  Therefore  this  solu- 
tion is  disregarded  and  3  and  2  are  the  required  values  of 
x  and  y,  respectively. 


238 


Second-Year  Mathematics 


250.  From  the  preceding  solution  it  is  seen  that  a  system 
of  equations  in  two  unknowns,  when  one  of  the  equations  is 
of  the  first  and  the  other  of  the  second  degree,  may  be  solved 
as  follows: 

Solve  the  linear  -equation  for  one  of  the  unknowns,  x  or  y, 
and  substitute  that  value  in  the  second-degree  equation.  This 
will  lead  to  a  second-degree  equation  in  the  other  unknown, 
y  or  x,  as  (5),  which  is  then  to  be  solved.  The  values  ofy  or  x, 
thus  found,  may  then  be  substituted  in  the  first-degree  equation, 
as  (i),  to  determine  the  corresponding  values  of  the  other 
unknown. 

This  method  of  solving  a  system  of  equations  in  two 
unknowns  is  called  "elimination  by  substitution." 

EXERCISES 

1.  Construct  a  right  triangle  whose  perimeter  is  30  and 
whose  hypotenuse  is  13. 

2.  In  the  right  triangle  ABC  (Fig.  285),  the  perimeters 
of  A  C  D  and  BCD  are  equal.     C  B  =4,  and  D  B  =  2.     Find 
A  C  and  A  D. 


D 

FIG.  285 


3.  In  the  right  triangle  of  Fig.  286,  with  sides  53;,  3^,  and 
x+y  =  5-     Construct  the  triangle. 

4.  Solve  x*+y3  =  2$ 

y—x=i. 

5.  Solve 


Areas  of  Polygons 


239 


6.  Solve 


Quadratic  Equations  Solved  by  the  Graph 

251.  Problems  which  lead  to  quadratic  equations  in  two 
unknowns  may  be  solved  by  means  of  the  graph. 

i.  Solve  x3+y'  =  2$  and  y—  x=i  by  the  graph.  By  as- 
suming values  for  x,  and  solving  x3+y3  =  2$  for  y,  we  have 
the  following  solutions  of  the  equation  x2+y3  =  2$: 


y=±5    y= 


y=±4  y= 


'3.4 


FIG.  287 

Plotting  these  solutions  (Fig.  287)  we  find  that  the  graph 
of  x2+yx  =  2$  is  a  circle  whose  center  is  at  the  origin  and 
whose  radius  is  1/25,  or  5. 

The  equation  xa+y*  =  2$  expresses  the  fact  that  the  sum 
of  the  squares  of  the  co-ordinates  of  any  point  on  the  graph 


240 


Second-  Year  Mathematics 


of  the  equation  is  25,  e.g.,  O  ~P1*=x2+y2  =  2$.  Hence 
O  P  =  5-  Moreover,  a  line  every  point  of  which  has  the  same 
distance  from  a  given  point  is  a  circle.  Therefore  the  graph 
of  x2+y2  =  2$  is  a  circle  whose  radius  is  5. 

The  points  of  intersection  of  this  circle  and  the  straight 
line  graph  of  equation  y— x  =  i,  are  Pj(3,  4),  and  P2(— 4,  —3). 

;=3 


j*=; 
/  y=' 


and 


Thus 
and  y. 

2.  In  triangle  ABC  (Fig.  2$ 
C=9o°.     Construct  the  triangle. 


are  the  required  values  of  x 
18),  AB=5,  CD—V-,  angle 


D 


J3 


FIG.  288 


FIG.  289 


3.  The  perimeter  of  the  rectangle  (Fig.  289)  is  34.     Find 
the  dimensions. 

4.  Solve   by  eliminating   by  substitution,   and  verify  by 

,    r  graphing: 

(1)  x2=s8-y2 
y  —  io—y 

(2)  xa+ya=4o 
x  — 3;y=o. 

5.  In  triangle  ABC  (Fig. 
290)  draw  D  E  parallel  to  A  B 


FIG.  290 


so  that  D  E  is  the  mean  pro- 


portional between  A  C  and  D  C. 

6.  Rectangles  R,  and  R2  have  equal  perimeters  (Fig.  291), 
/?!  and  R3  have  equal  areas.     Find  the  dimensions  of  Rt. 


Areas  of  Polygons 


241 


7.  In  the  right  triangle  ABC  (Fig.  292),  find  the  lengths 
of  the  sides  including  the  right  angle.     Area=2. 


y 

FIG.  291 

w2+«2  =  i7     Why? 
m  •  n=4       Why? 
Adding  2  times  (2)  to  (i) 


Subtracting  2  times  (2)  from  (i): 
m2  —  2  mn  +  n2  =  9 


(i) 

(2) 


(3) 


(4) 


FIG.  292 

Thus  the  problem  of  solving  (i)  and  (2)  is  reduced  to 
solving  (3)  and  (4),  i.  e.,  solving  the  following  system  of 
equations: 

=~s    (5) 


w—  »= 


w—  w= 


m—  n=—  3 


w—  n=—  3 

8.  Solve  Problem  7  by  the  graph. 

Equation  (i)  of  Problem  7  is  a  circle  whose  radius  is  -\/rj. 
Equation  (2)  expresses  the  fact  that  the  rectangle  formed  by 
the  co-ordinates  of  any  point  on  its  graph  is  constant  and 
equal  to  4.  This  curve  is  called  the  rectangular  hyperbola 
(see  Fig.  293).  Find  the  points  of  intersection  of  the  two  curves 
and  determine  their  co-ordinates. 


242 


Second-  Year  Mathematics 


9.  Plot  on  the  same  figure  equations  (3)  and  (4).  Show 
from  the  graph  why  the  solutions  of  the  systems  (5)  are  the 
same  as  the  solution  of  (i)  and  (2). 


-8 


0 


-6 


I  FIG.  293 

10.  Solve  for  m  and  n  the  following  equations: 
w2+w2  — 13=0 
mn— 6=0 

u.  Verify  by  graphing  the  solutions  of  Problem  10. 

12.  Solve  x2-\-y2=a  and  xy  =  b. 

13.  In  the  right  triangle  ABC  (Fig.  294),  x  and  y  have 


243 


the  same  values  as  in  the  right  triangle  D  E  F.     Find  the 
values  of  x  and  y. 

Suggestion:    Eliminate  x1  or  y3  as  in  the  case  of  a  linear  equation  in 
x  and  y. 


D 

FIG.  294 

14.  Verify,  by  graphing,  the  solutions  of  Problem  13. 

15.  Solve  x2+y2  =  a,  and  x2—  y2=b. 

PROBLEMS  AND  EXERCISES  FOR  REVIEW 

1.  Draw  through  a  vertex  of  a  triangle  lines  dividing  it 
(i)  into  two  parts  one  of  which  shall  be  (a)  §,  (6)  £,  (c)  ;}  of 
the  other;   (2)  into  parts  in  the  ratio  of  2  :  3  :  4. 

2.  Draw  a  parallel  to  one  side  of  a  triangle  cutting  off  a 
triangle  which  shall  be  (i)  £,  (2)  ?,  (3)  £  of  the  given  triangle. 


M 


FIG.  295 

3.  To  bisect  a  triangle  by  a  line  through  a  given  point  P 
on  the  perimeter  not  the  vertex  of  an  angle  (see  Fig.  295). 

Draw  median  B  M,  also  P  M,  B  D  ||  P  M,  and  P  D.  A?  M  D 
-  A?  M  B  (  ?).  . '.  A  A  D  P  =  quadrilateral  D  P  B  C.  (?) 


244  Second-Year  Mathematics 

4.  To  trisect  a  triangle  by  lines  through  a  point  P  on  one 
side  not  at  the  vertex. 

Use  the  method  of  Exercise  3. 

5.  Construct  (i)  a  right  triangle,  (2)  an  isosceles  triangle, 
(3)  an  obtuse-angled  triangle  each  equal  to  a  given  triangle. 

6.  Construct  an  equilateral  triangle  equivalent  to  a  given 
triangle. 

Transform  the  given  triangle  into  an  equal  triangle  having  one 
angle  60°.  Apply  the  theorem — two  triangles  having  an  angle  in  each 
equal  are  to  each  other  as  the  products  of  the  sides  including  the  equal 
angles  (see  §  248,  p.  231). 

7.  On  a  given  base,  construct  a  triangle  equal  to  a  given 
triangle  not  having  a  side  equal  to  the  given  base. 

8.  Find  a  line  equivalent  to  the  ]/6  (see  p.  236,  Problem  6). 

9.  Construct  a  triangle  equivalent  to  a  given  hexagon. 

10.  Construct  a  circle  equivalent  to  6  times  a  given  circle 
(see  Exercise  8). 

11.  Construct  a  pentagon  equal  to  -f-  of  a  given  pentagon. 

12.  A  line  parallel  to  the  base  of  a  triangle  cuts  off  a  tri- 
angle equal  to  f  of  it.     If  one  side  of  the  triangle  is  12,  how 
far  from  the  vertex  does  the  line  cut  it  ? 

13.  The  diagonals  of  a  rhombus  are  2^—14  and  2X,  and 
a  side  is  x+i.     Find  x. 

14.  The  radii  of  two  circles  are  25  and  24.     Find  the 
radius  of  a  circle  equivalent  to  their  difference.     Use  the  law  of 
similar  figures. 

15.  The  area  of  one  of  three  circles  is  equal  to  the  sum 
of  the  other  two,  and  their  radii  are  x,  x— 7,  x+i.     Find  x. 

1 6.  The   difference   of  two   circles  whose  diameters  are 
x+2  and  x  is  equivalent  to  a  circle  whose  diameter  is  x—j. 
Find  x. 

17.  The  area  of  a  rectangle  is  60  and  its  diagonal  is  13. 
Find  its  dimensions. 


Areas  of  Polygons  245 

1 8.  The  perimeter  of  a  rectangle  is  46  and  its  area  is  120. 
Find  its  dimensions. 

19.  The  perimeter  of  a  rectangle  is  62  and  its  diagonal  is 
25.     Find  its  area. 

20.  The  sides  of  a  triangle  are  17,  10,  and  9.     The  altitude 
of  a  similar  triangle  upon  the  side  homologous  to  the  side  10 
in  the  given  triangle  is  14^.     Find  all  the  sides  of  the  second 
triangle. 

21.  Find  the  side  of  a  square  equivalent  to  an  equilateral 
triangle  whose  side  is  4^/3. 

22.  The  ratio  of  two  similar  polygons  is  (i)  9  :  4,  (2)  2  :  5, 
and  the  shortest  side  of  the  first  is  12.     Find  the  shortest  side 
of  the  second. 

23.  The  base  of  a  triangle  is  18  feet.     Find  the  length  of 
a  line  parallel  to  the  base  which  bisects  the  triangle. 

24.  The  side  of  a  square  (or  of  any  polygon,  or  the  radius 
of  a  circle)  is  a.     Find  the  side  (or  radius)  of  a  similar  figure 
K  times  as  large. 

25.  Compute   the   altitude  upon   the   hypotenuse   of   the 
right  triangle  A  B  C  in  terms  of  its  legs  a  and  b. 

26.  The  altitude  and  base  of  a  rectangle  are  in  the  ratio 
of  8  to  15  and  the  diagonal  is  34  ft.     Find  the  area. 

27.  Bisect  any  parallelogram  by  a  line  drawn  through  any 
point  on  its  perimeter. 

28.  To  transform  any  triangle,  ABC,  into  an  equivalent 
right  triangle  containing  one  of  the  acute  angles. 

29.  The  dimensions  of  a  rectangle  are  in  the  ratio  of 
2ab  to  a2—b2,  and  the  diagonal  is  a2c2+&2c2.     Find  the  area. 

2d 

30.  The  ratio  of  the  sides  of  a  rectangle  is  —    —  ,  and  the 

a2-i 

diagonal  is  6a2+  6.     Find  the  sides. 


CHAPTER  VII 


REGULAR  POLYGONS  INSCRIBED  IN,  AND  CIRCUMSCRIBED 
ABOUT,  A  CIRCLE 

252.  A  polygon  that  is  equilateral  and  equiangular  is  a 
regular  polygon.     Show  that  an  equilateral  triangle  is  a  regular 
polygon;   that  a  square  is  a  regular  polygon. 

A  polygon  whose  vertices  lie  on  a  circle  and  whose  sides 
are  chords  is  an  inscribed  polygon.  The  circle  is  said  to  be 
circumscribed  about  the  polygon. 

A  polygon  whose  sides  are  tangent  to  a  circle  is  a  circum- 
scribed polygon.  The  circle  is  said  to  be  inscribed  in  the 
polygon. 

To  determine  whether  inscribed  or  circumscribed  polygons 
are  regular,  the  following  two  theorems  may  be  used: 

PROPOSITION  I 

253.  Theorem:    If  a  circle  is  divided  into  equal  parts  and 
if  the  successive  points  of  division  are  connected  by  line-segments, 
the  polygon  so  formed  is  a  regular  inscribed  polygon. 

Hypothesis:       A~B=B~C  =  CfD,  etc.  (Fig.  296); 

C 


Conclusion : 


(     (a)  AB=BC  =  CD,  etc. 

inn  •   / 

246 


Regular  Polygons  in  and  about  a  Circle 


247 


Prove  (&)  by  the  method  of  congruent  triangles,  using 
AsB  AF,  ABC,  BCD,  etc. 

PROPOSITION  II 

254.  Theorem:  If  a  circle  is  divided  into  equal  parts,  and 
if  tangents  are  drawn  at  the  points  of  division,  a  regular  cir- 
cumscribed polygon  is  formed. 

Prove  AM  A  L^  ALB  N^ANCO,  etc.  (Fig.  297). 


FIG.  297 

Then  Z.A  =  /_B  =  Z.C,  etc.,  the  first  condition  for  a 
regular  polygon. 

MA=AL=LB=BN=NC,  etc.     Why? 

AL=BN  and  LB=NC.     Why? 

.-.  AB=BC. 

In  a  similar  way,  prove  B  C  =  C  D,  etc.,  the  second  condi- 
tion for  a  regular  polygon. 

EXERCISES 

1.  Prove   that  an  equiangular  circumscribed  polygon   is 
regular. 

2.  Prove  that  an  equilateral  polygon  inscribed  in  a  circle 
is  a  regular  polygon. 

Use  §  253. 


248 


Second-Year  Mathematics 


PROPOSITION  III 

255.  Problem:   To  inscribe  a  square  in  a  given  circle. 
Construction :     Draw   the   diameter   A  B  _l_  diameter    C  D 

(Fig.  298).    Join  A  to  D,  D  to  B,  etc.    To  prove  that  A  D  B  C 

is  a  square,  Proposition  I  may  be  used. 

A 


PROPOSITION  IV 

256.  Problem:    To  circumscribe  a  square  about  a  circle. 
Construction:    In  Fig.  298,  draw  tangents  at  A,  C,  B,  D. 
The  proof  follows  from  Proposition  II. 

EXERCISES 

1.  Denoting  the  side  of  the  inscribed  square  by  x,  and  the 
radius  of  the  circle  by  r,  prove  that  x=r~\/2. 

2.  Express  the  side  of  the  circumscribed  square  in  terms 
of  the  radius  of  the  circle. 

3.  Express  in  terms  of  the  radius,  the  areas  of  the  inscribed 
and  circumscribed  squares. 

4.  Express  the  perimeters  of  the  inscribed  and  circum- 
scribed squares  in  terms  of  the  radius;  in  terms  of  the  diameter. 

5.  The  radii  of  two  circles  are  6  cm.  and  i  cm.,  respectively. 
Find  the  ratio  of  the  areas  of  the  inscribed  squares;    of  the 
circumscribed   squares.     How   does   the   ratio   of   the    areas 
compare  with  the  ratic  of  the  squares  of  the  radii  ? 


Regular  Polygons  in  and  about  a  Circle 


249 


6.  The  area  of  a  square  is  16  sq.  cm.     Find  the  diameters 
of  the  inscribed  and  circumscribed  circles. 

7.  Prove  that  the  point  of  intersection  of  the  diagonals  of 
a  square  is  the  center  of  the  inscribed  and  circumscribed 
circles. 

8.  Show  how  to  construct  a  regular  8-side,  i6-side,  32-side. 

PROPOSITION  V 

257.  Problem:    To  inscribe  a  regiilar  hexagon  in  a  circle. 
Construction:   With  A   as   center 

and  radius  O  A  (Fig.  299),  strike  an 
arc  meeting  the  circle  at  B. 

With  B  as  center  and  the  same 
radius,  strike  an  arc  at  C,  D,  etc. 
Draw  the  chords  A  B,  B  C,  C  D,  etc. 

Then  A  B  C  D  E  F  is  the  required 
hexagon. 

Proof:   Draw  O  B. 

Prove    ZO=6o°=£of  360°. 

A~B  =  £  of  the  circle.     Why  ? 
B~C = £  of  the  circle.     Why  ? 

ABCDEFisa  regular  hexagon.     Why  ? 

PROPOSITION  VI 

258.  Problem:     To  inscribe  an  equilateral  triangle  in  a 
circle. 

Solve  the  problem  and  give  proof. 

EXERCISES 

1.  Inscribe  a  regular  i2-side,  24-side,  48-side,  etc. 

2.  Given  the  radius  r  of  a  circle.     Express  in  terms  of  r 
the  side  of  the  regular  inscribed  hexagon  (Fig.  300). 


FIG.  299 


250 


Second-Year  Mathematics 


3.  Given  the  radius  r  of  a  circle,  express  in  terms  of  r  the 
side  of  the  regular  inscribed  triangle. 

Bisect  angle  ACB   (Fig.   301).     .'.  x  =  \  •  36o°  =  i2o°,   /.  y  =  6o°. 

AE  I  CD  (why?),  .'.  A  E2=r*--(why ?),  .'.  AB=r1/3  (why?). 

4 


4.  Prove  that  the  area  of  the  equilateral  inscribed  triangle 


s 


5.  To  circumscribe  about  a  circle  a  regular  hexagon,  an 
equilateral  triangle,  a  regular  i2-side,  24-side,  etc  ..... 

6.  Prove  that  the   side  of  the  circumscribed  equilateral 
triangle  is  2r]/3  and  the  area  3^2v/3- 

7.  Prove  that  the  side  of  the  circumscribed  regular  hexagon 
is  §n/3~,  and  the  area  2/-2j/3. 

8.  Express  in  terms  of  the  radius:  the  perimeters  (a)  of  the 
inscribed   and  circumscribed   regular   hexagons;     (b)    of   the 
equilateral  inscribed  and  circumscribed  triangles. 

9.  Find  the  ratio  of  the  inscribed  and  circumscribed  equi- 
lateral triangles. 

10.  Find  the  distance  from  the  center  of  a  circle  to  the 
side  of  the  equilateral  inscribed  triangle  (the  apothem). 

11.  Find  the  area  of  a  regular  hexagon  whose   side  is 
6  inches. 

12.  The  radius  of  a  circle  is  10.     Find  the  area  of  the 
inscribed  regular  hexagon. 


Regular  Polygons  in  and  about  a  Circle 


251 


13.  The  diameter  of  a  circle  is  8.     Find  the  area  of  the 
regular  inscribed  hexagon. 

14.  Prove  that  in  the  same  circle  the  area  of  a  regular 
inscribed  hexagon  is  twice  as  large  as  that  of  the  inscribed 
equilateral  triangle. 

259.  In  order  to  construct  a  regular  lo-side    (decagon) 
in  a  circle,  it  is  necessary  to  divide  a  line-segment  in  extreme 
and  mean  ratio. 

A  line-segment  is  divided  in  extreme  and  mean  ratio  when 
the  longer  part  is  a  mean  proportional  between  the  whole 
segment  and  the  shorter  part.  Thus  a  line-segment  A  B  is 
divided  in  extreme  and  mean  ratio  at  C,  when  A  B  :  A  C  = 
A  C  :  B  C. 

260.  Problem:     To  divide  a  line-segment  in  extreme  and 
mean  ratio. 

Let  A  B  be  the  given  line- 
segment  (Fig.  302). 

To  find  the  point  C,  such 
that  A  B  :  AC=AC  :  B  C. 

Construction:  Draw  B  F 
J.AB,  and  =  £A  B. 

With  F  as  center  and 
radius  F  B  draw  0  F. 


FIG.  302 


Draw  A  F  cutting  the  circle  at  E  and  D. 
Lay  off  on  A  B,  A  C=A  E.     C  is  the  required  point. 
AD    AB 


Proof: 


AB     AE 
AD-AB 


(?)  (§182,  p.  150) 
AB-AE 


(i) 


AB  AE        (?)  (Exercise  3,  p.  80).    (2) 

Substituting  equal  values  in  (2),  noting  that  A  B=E  D  (?) 


and  A  E=A  C  (?),  we  obtain  -  ^=-7-7;,  then  -r-= 

A  D     A  \^  A.  (^ 

Q.E.D. 


AC 
:BC 


252  Second-Year  Mathematics 

EXERCISES 

1.  Divide  a  line-segment  5-in.  long  in  extreme  and  mean 
ratio. 

Use  compasses. 

2.  Calculate  algebraically  the  length  of  the  longer  segment 
of  the  line-segment  in  Exercise  i;   of  one  6  in.  long;   of  one 
a  in.  long;  of  one  b  in.  long. 

PROPOSITION  VII 

261.  Problem:    To  inscribe  a  regular  lo-side  (decagon)  in 
a  circle. 

Given  ©  O. 

Required  to  inscribe  a  regular  decagon  (lo-side). 

Construction:    Divide  the  radius  O  A  (Fig.  303)  in  extreme 

and  mean  ratio  at  B  (see  §  260),  making  77^=^-7-  • 

(J  x>     x>  A 

With  A  as  center  and  O  B  as 
radius,  strike  an  arc  at  C. 

With  C  as  center  and  the  same 
radius,  strike  an  arc  at  D,  E, 
etc. 

Draw  AC,  CD,  D  E,  etc.  Then 
A  C  D  E  .  .  .  .  is  the  required 
decagon. 

Proof:   Draw  O  C  and  B  C.     Since  77^=^-7-  and  O  B  = 

(J  D       D  A 

OA    AC 

AC'thenAC=BA' 

and  as  Z^  =  Z^,  then  AAOC^ABCA  (why?). 
OA    CB  OA     i  CB     i 

Then  OC  =  CA  (why?)'  OC=?       '  = 

andCB=CA. 


Regular  Polygons  in  and  about  a  Circle  253 

Therefore  m=n.    Why? 

Prove:  p=o,  and  n  =  2  •  o,  m  =  2  •  o,  q+p  =  2>o 

o+w+<7+/>  =  i8o     (Why?) 

0  +  20  +  20  =  180     (Why?) 

.'.  0=36°. 

Therefore  A  C=T^  of  circle. 
.'.  A  C  D  E  .  .  .  .  is  a  regular  inscribed  lo-side.     Q.E.D. 

PROBLEMS 

1.  To  inscribe  a  regular  5-side  (pentagon)  in  a  given  circle. 

2.  To  circumscribe  a  regular  ic-side  about  a  given  circle. 

3.  To    circumscribe    a    regular    5-side,    2o-side,    4o-side, 
etc about  a  given  circle. 

The  Side  of  a  Regular  Inscribed  Decagon 
262.  Given  the  radius  r  of  a  circle,  and  s  the  side  of  the 
decagon.     Compute  the  side  of  the  regular  inscribed  decagon 
(see  Fig.  304). 

(Why?)     /.  r*-rs=s' 


(?) 


Show  that  only  the  positive  sign  before  FIG.  304 

the  radical  can  be  used. 

The  Side  of  a  Regular  Inscribed  Pentagon 
263.  Show  that  the  side  of  a  regular  inscribed  pentagon  is 

equal  to  ^rl/io  —  21/5. 

In  Fig.  305  let  A  C  and  C  B  be  sides  of  a  regular  decagon, 

then  A  B  is  the  side  of  a  regular  pentagon.    Why  ? 

In  right  triangle  AFC,  A~F2  =AC2  -FC2  (i) 


254 


Second-Year  Mathematics 


Use  s  for  A  F, 
for  since  OF2=r2-s2  (from  AAFO), 


"1)  f°r  A  C,  and  r-vV-s2  for  F  C, 
=  l/r2-i2. 


Substituting  these  values  in  (i) 


Whence  $=-l/io  —  21/5. 
4 


(3) 


But  A  B  is  2s,  hence  A  B,  the  side  of  the  regular  pentagon, 

.    r  , — 

is  -V  10  —  21/5. 

2 

EXERCISES 

1.  Find  an  approximate  value  of  (1/5  —  1)  to  be  1.236  +  . 

2.  Find  an  approximate  value  of  1/10  —  21/5  to  be  2 . 351  + 

3.  Using  the  approximate  value  in  Exercise  2,  find  the  side 
of  a  pentagon  inscribed  in  a  circle  of  radius  8;   10;   15;  a. 

4.  Using  the  approximate  value  in  Exercise  i,  find  the  side 
of  a  decagon  inscribed  in  a  circle  of 'radius  8;  10;  15;  a. 

5.  The  side  of  an  inscribed  pentagon  is  18.8  in.    Find  the 
radius  of  the  circumscribed  circle. 

6.  The  side  of  an  inscribed  decagon  is  14.83  in.     Find  the 
radius  of  the  circumscribed  circle. 

7.  A  man  has  a  round  table  top  which  he  wishes  to  change 
into  a  pentagon.     The  diameter  of  the  top  is  2^  ft.     What  is 
the  length  of  the  cut  required  ? 


Regular  Polygons  in  and  about  a  Circle 


255 


PROPOSITION  VIII 

264.  Problem:  To  construct  a  triangle  whose  sides  shall 
be,  respectively,  the  side  of  a  regular  hexagon,  the  side  of  a  regular 
decagon,  and  the  side  of  a  regular  pentagon,  inscribed  in  a 
given  circle. 

Given  0  O  (Fig.  306). 

Draw  diameter  A  B,  and  radius 
OC_LAB. 

Bisect  O  B  at  M. 

With  M  C  as  radius  and  M  as 
center,  draw  arc  C  E,  cutting  A  B 
at  E.  Join  C  and  E. 

Prove  E  O  C  to  be  the  required 
triangle : 

(a)  C  O  to  be  the  side  of  a  hexagon  (radius).  (a) 

(6)  E  O  to  be  the  side  of  a  decagon  [$r(j/5  —  *)]•  See  §  262. 

(c)  EC   to   be   the   side  of   a  pentagon  [£r I/ 10  — 
See  §263. 


FIG.  306 


Proof:    In  right  triangle  C  O  M,  C  M  =-+r2,    .'.  C  M  = 

4 

p/5.     Why? 

EO  =  CM-OM  =  -i/5--=-(l/5-i),    the    side   of    a 
decagon.  (b) 

In    right    triangle    E  O  C,     EC*-r*+|-(>/s-i)T- 

-l/io— 2 j/ 5,  the  side  of  a  pentagon.  (c) 


EXERCISES 
i.  With  radius  equal    to   line-sect 


_o ,  construct  a 


triangle  with  sides  equal  to  the  regular  hexagon,  decagon,  and 
pentagon  by  Proposition  VIII. 


256 


Second-Year  Mathematics 


With  compasses  test  whether  the  line  E  O,  used  as  a 
chord,  cuts  the  circle  into  ten  equal  arcs,  and  E  C,  into  five 
equal  arcs. 

2.  The  radius  of  a  circle  is  8  in.     Find  the  side  of  a  regular 
inscribed  decagon. 

3.  The  radius  of  a  circle  is  10  in.     Find  the  side  of  a 
regular  inscribed  pentagon. 

4.  With  radius  =  7  in.,  find  the  side  of  the  regular  inscribed 
decagon. 

5.  With  radius  =  7  in.,  find  the  side  of  the  regular  inscribed 
pentagon. 

6.  The  side  of  a  regular  pentagon  is  5  in.     Find  the  radius 
of  the  circumscribed  circle. 

7.  The  side  of  a  regular  decagon  is  10  in.     Find  the  radius 
of  the  circumscribed  circle. 

8.  What  part  of  the  area  of  a  circle  is  the  area  of  the  tri- 
angle drawn  in  it  by  Proposition  VIII,  the  area  of  the  circle 
being  -n-R3  ? 

Use    ir=  —  . 


265.  Problem: 

Since  15°  is 


PROPOSITION  IX 
To  inscribe  a  regular  ij-side. 
of  360°,  the  required  polygon  may  be 
obtained    by   constructing    an    angle   of 
24°  with  the  vertex  at  O.     Notice  that 
24°  =60°  -36°. 

Construct  A  B,  the  side  of  a  regular 
hexagon  (Fig.  307),  making  angle 
AOB=6o°. 

Then  construct   A  C,   the  side    of   a 
FIG.  307 

regular  decagon,  making  A  O  0  =  36°. 
Then  B  O  C  =  24°,  and  B~C  is  TV  of  the  circle. 


Regular  Polygons  in  and  about  a  Circle 


257 


EXERCISES 

1.  Inscribe  a  regular  3o-side,  6o-side,  etc 

2.  To  circumscribe  a  regular  polygon  of  i5-sides. 

3.  Show  that  regular  polygons  can  be  inscribed  in  a  circle, 
having   the   following   number   of   sides:     2",    3  •  2",    5  •  2", 
15  •  2". 

Exercise  3  shows  that  a  circle  can  be  divided  into  2,  3  •  2, 
5  •  2,  15  •  2  equal  parts. 

Gauss  proved  that  by  the  use  of  a  compass  and  unmarked 
straight  edge,  a  circle  could  be  divided  into  2>f  +  i  equal  parts 
where  n  must  be  a  number  making  2n  +  i  a  prime  number. 
Explain. 

Circles  Circumscribed  about,  and  Inscribed  in,  a  Regular  Polygon 
PROPOSITION  X 

266.  Theorem:  A  circle  can  be  circumscribed  about  any 
regular  polygon. 

Let  A  B  C  D  .  .  .  .  be  a  regular 
polygon  (Fig.  308). 

To  construct  a  circle  circum- 
scribed about  A  B  C  D 

Construction:  Draw  a  circle 
through  three  vertices,  as  A,  B, 
and  C  (?).  This  is  the  required 
circle. 

Proof:    We  are  to  prove    that  FIG.  308 

circle  ABC  passes  through  D,  E,  etc. 

Prove  A  A  O  B^  AB  O  C. 

.'.   y=u,  z=u  (?),  y+z  =  u+p  (?). 

Therefore,  u+u—u+p,  and  u=p. 

Prove  AGO  D^  AC  OB.    Then  O  D  =  O  B  (?). 

Circle  ABC  passes  through  D.     ( ?) 


258 


Second-  Year  Mathematics 


In  a  similar  way  circle  ABC  may  be  proved  to  pass  through 
E,  etc.  Q.E.D. 

PROPOSITION  XI 

267.  Theorem:    A  circle  can  be  inscribed  in  any  regular 
polygon. 

Let  A  B  C  ....  be  a  regular  polygon  (Fig.  309). 

Required  to  draw  the  inscribed 
circle. 

Construction :  Construct  as  in  Pro- 
position X  the  center  O  of  the  circum- 
scribed circle. 

From  O  draw  OKiAB.  With 
O  as  center  and  radius  O  K  draw 
circle  H  K  I.  This  is  the  required 
circle. 

Proof:  Draw  the  circle  circumscribed  about  ABC.... 
(§266).  DrawOL^AE,  OM_LB  C.  O  K  =  O  L  =  O  M, 
etc.  Why  ? 

Since  O  K JL  A  B,     .'.  A  B  is  tangent  to  H  K  I.     (?) 
Similarly  prove  that  B  C,  C  D,  etc.,  are  tangents  to  H  K  I. 

Therefore  H  K  I  is  inscribed  in  polygon  ABC (?) 

Q.E.D. 

To  Find  the  Circumference  of  a  Circle 

268.  A  regular  polygon  may  be  inscribed  in  a  circle,  by 
dividing  the  circle  into  equal  parts  and  joining  the  successive 
points  of  division  by  segments.     (See 

Proposition  I,  p.  246.) 

269.  A  regular  polygon  may  be  cir- 
cumscribed about  a  circle  by  drawing 
tangents  at  these  points  of  division  of 
the  circle.     (See  Proposition  II,  p.  247.) 

270.  A  regular  polygon  may  also  FIG.  310 


Regular  Polygons  in  and  about  a  Circle 


259 


be  circumscribed  about  a  circle  by  drawing  tangents  parallel  to 
the  sides  of  the  inscribed  polygon,  as  in  Fig.  310. 

To  prove  this,  we  will  first  prove  the  following  theorem: 

PROPOSITION  XII 

271.  Theorem:  If  at  the  midpoint  of  the  arcs  subtended  by 
the  sides  of  a  given  regular  inscribed  polygon,  tangents  are  drawn 
to  the  circle,  they  are  parallel  to  the  sides  of  the  given  polygon 
and  form  a  regular  circumscribed  polygon. 

To  prove  that  A  B  ||  A'B'  (Fig.  311),  show  that  both  A  B 
and  A'B'  are  perpendicular  to  O  P. 


To  prove  that  C'A'B'D'  ....  is  regular,  show  that  arcs 
Q  P,  P  R,  R  S,  etc.,  are  equal. 

EXERCISES 

1.  In  Fig.  311,  prove  that  line  OB  passes  through  B', 
and  state  this  exercise  as  a  theorem. 

Prove  that  B  O  bisects  angle  P  O  R,  and  therefore  contains  point  B'. 

2.  Denoting  the  side  of  the  circumscribed  regular  polygon 
of  w-sides,  by  Sn  and  of  the  regular  inscribed  polygon  of  the 

same  number  of  sides  by  sn,  prove  that  Sn=—^=     "          . 


260 


Second-  Year  Mathematics 


CD     OF 

Suggestion:    In  Fig.  312 


Therefore  ^  =  V  r*-(ls  )>    twhy?) 


Solve  this  equation  for  Sn- 

3.  Taking  the  value  of  s3  from  §  258,  Exercise  3,  find  by 
means  of  the  formula  of  Exercise  2  the  value  of  S3. 

C  F  D 


FIG.  312 

4.  In  a  similar  way  find  the  value  of  S4,  using  s4=r}/2, 
as  proved  in  §  256,  Exercise  i. 

5.  Find  Sf,,  using  56  =r. 

PROPOSITION  XIII 

272.  Theorem:    If  a  regular  polygon  of  n-sides  is  inscribed 
in  a  circle  and  the  arcs  subtended  by  the  sides  are  bisected,  a 
regular  polygon  of  2n-sides  may  be  drawn  by  joining  each  point 
of  division  to  the  adjacent  corners  of  the  given  polygon.     Prove. 

PROPOSITION  XIV 

273.  Theorem:    The  perimeter  of  a  regular  inscribed  2n- 
side  is  greater  than  the  perimeter  of  the  regular  n-side  inscribed 
in  the  same  circle.     Prove. 

PROPOSITION  XV 

274.  Theorem:    The  perimeter  of  a  regular  circumscribed 
2ii-gon  is  less  than  the  perimeter  of  the  regular  n-gon  circum- 
scribed about  the  same  circle.     Prove. 


Regular  Polygons  in  and  about  a  Circle  261 

275.  From  Propositions  XIV  and  XV  it  follows  that  the 
perimeter  of  the  regular  inscribed  polygon  increases  as  the 
number  of  sides  increases: 

While  the  perimeter  of  the  regular  circumscribed  polygon 
decreases  as  the  number  of  sides  increases. 

In  the  following  discussion  it  will  be  shown  that  by  in- 
creasing the  number  of  sides  of  the  inscribed  and  circumscribed 
regular  polygons,  the  perimeters  approach  each  other  more 
and  more,  and  that  the  decimal  fractions  expressing  these  two 
perimeters  can  be  made  to  agree  to  a  greater  and  greater 
number  of  decimal  places.  It  will  be  assumed  that  the  cir- 
cumference of  the  circle  is  less  than  the  perimeter  of  any 
polygon  circumscribed  about  the  circle,  and  it  is  easily  proved 
that  the  circumference  is  greater  than  the  perimeter  of  any 
polygon  inscribed  in  the  circle.  Hence,  the  length  of  the  cir- 
cumference lies  between  the  lengths  of  the  perimeters  of  any 
two  inscribed  and  circumscribed  polygons. 

To  calculate  these  lengths,  the  two  following  propositions 
give  formulas  for  determining  the  perimeters  of  circumscribed 
and  of  inscribed  polygons  of  a  greater  and  greater  number  of 
sides. 

PROPOSITION  XVI 

276.  Problem:    To  compute  the  side  of  the  regular  circum- 
scribed 2n-gon  in  terms  of  the  sides  of  the  regular  inscribed  and 
circumscribed  n-gon. 

Let  A  B  be  the  side  of  the  regular  inscribed  w-gon,  denoted 
by  sn  (Fig.  313). 

Let  C  D  be  the  side  of  the  regular  circumscribed  2«-gon, 
denoted  by  S2n.  E  G  the  side  of  the  regular  circumscribed 
w-gon  denoted  by  Sn. 

To  find  S2n  in  terms  of  sn  and  Sn. 
CD    CG 


262 


Second-Year  Mathematics 
S      S 


n          A  G  S_n  Sn 

2 

S       S  —S 
Therefore  -—  =   "  „ "  —  ,  and  S2n  •  Sn=sn(Sn—S2n). 


Or,  52W  = 


FIG.  313 

EXERCISE.     Denoting  the  perimeters  of  the  inscribed  and 
circumscribed  polygons  by  P2n,  pn,  Pn,  prove  that 

j.          n 

(2) 


PROPOSITION  XVII 

277.  Problem:  Find  the  side  of  the  regular  inscribed  poly- 
gon of  2n-sides  in  terms  of  the  side  of  the  regular  circumscribed 
polygon  of  2n-sides,  and  of  the  side  0}  the  regular  inscribed 
polygon  of  n-sides. 

Let    A  B  be  denoted  by  sn  (Fig.  314). 
A  H  be  denoted  by  s2n. 
C  D  be  denoted  by  S2n. 
To  find  s2n  in  terms  of  sn  and  S2n. 


Proof:    ACHI^AHAK.     Hence, 


~      - 
AH     AK 


(Why?) 
' 


Regular  Polygons  in  and  about  a  Circle  263 


Whence,  ~  =—  .     (Why  ?) 


Therefore,   S2n  •  —=s2n  •  s2n.     (Why?) 


and  s2n=- 


V--7- 


FIG.  314 


(i) 


EXERCISES 


1.  Prove  that  p2n=vP2n  •  pn.  (2) 

2.  It  was  proved  in  §256,  Exercise  i,  that  s4=/-j/2,  and 
in  §  256,  Exercise  2,  that  S4—d,  the  diameter  of  the  circle. 

Prove  that  P4—^dt  and  p4=dX  2. 828427 

3.  By  means   of   the  formula   (2)  of   §   276,   prove    that 
P&=dX 3. 313780  .  .  .  .  ,  and  by  formula  (2),  Exercise  i,  of 
this  list,  that  p&  =dX 3. 061467 

4.  It  has  been  proved  that  P6=dX3-464ioi   ....  and 
p6  =  $d.     (P.  250,  8.) 

Prove  that/*, 2=^X3. 2 1 5390  .  .  .  .  and pl3=dX$.  105828. 


264 


Second-  Year  Mathematics 


5.  Prove  the  following: 


Pi  =2-598  .  . 

.  .  d 

P3  =5.196  .  . 

.  .  d 

P4  =2.828  .  . 

.  .  d 

P4  =4.000  .  . 

.  .  d 

p6  =3.000  .  . 

.  .  d 

P6  =3.464  -  - 

.  .  d 

ps  =3.061  .  . 

.  .  d 

PS  =3-3i4  •  - 

.  .  d 

£I2  =  3.106  .   . 

.  .  d 

PI2=3.2i5  .  . 

.  .  d 

£i6=3-121    •    • 

.  d 

Pl6  =  3.i83  . 

.  d 

P.  = 


The  table  above  shows  how  the  decimal  fractions  expressing 
the  perimeters  agree  more  and  more  closely  as  the  number  of 
sides  of  the  polygon  is  increased 

278.  The  following  table,  which  gives  the  decimal  frac- 
tions to  six  places,  shows  the  approach  of  the  perimeters  still 
better: 

=  2.828427 
=3.000000 
=3.061467 
=3.105828 
Pl6  =3.182598  . 

P*4  =3-159659    ' 

P32  =3-I5l725    • 

P48  =3.146086    . 

P64  =3.144118    . 

Pp6  =3.142714    . 


=  4.000000 
=  3.464121 
=  3.313708 


d 


PI92=3.  141873 


pia 

pl6  =3.121445 
p24  =3.132623 
p32  =3.136548 


d 


£64    =3.140331    . 
£96    =3-I4I032    . 

£I28  =  3-i4i277  . 
£192=3-141452  - 
£2S6  =  3-I4i5I4  - 
£348  =  3 •I4i557  • 
The  last  two  perimeters  agree  to  three  decimal  places. 

Thus  the  circumference  of  the  circle  of  diameter  d,  toward 

which  these   perimeters  approach,  is  found  correct  to  three 

decimal  places.     It  equals  dX 3. 141 

Thus  as  the  perimeters  of  the  inscribed  and  circumscribed 

polygons  with  increasing  numbers  of  sides,  approach  each 


Regular  Polygons  in  and  about  a  Circle  265 

other  in  length,  both  of  them  approach  more  and  more  closely 
the  length  of  the  circumference  of  the  circle.  But  however 
close  the  length  of  the  perimeter  of  any  polygon  may  come  to 
the  length  of  the  circumference,  there  is  always  another  poly- 
gon the  perimeter  of  which  comes  still  closer  to  the  length  of 
the  circumference;  and  for  every  number  given  as  expressing 
the  difference  between  any  perimeter  and  the  circumference, 
we  can  find  a  polygon  whose  perimeter  differs  from  the  cir- 
cumference by  less  than  that  number.  This  is  expressed  by 
saying  that  the  perimeters  of  the  inscribed  and  circumscribed 
polygons  approach  the  circumference  as  a  limit. 

As  is  seen  by  the  table  above,  the  value  of  this  limit  can 
be  expressed  more  and  more  exactly  by  taking  polygons  of  a 
greater  and  greater  number  of  sides.  It  cannot,  however,  be 
determined  exactly. 

Continuing  to  increase  the  number  of  sides,  we  find  in 
the  table  above  ^8192=^X3.  1415928  ....  and  p$I92  = 


From  this  it  is  seen  that  the  circumference,  being  between 
P8lg2  and  pBi92,  can  be  expressed  by  0=^X3.141592  .... 
approximately,  with  an  error  less  than  i  millionth. 

The  circumference  of  the  circle  is  therefore  a  multiple  of 
the  diameter,  which  may  not  be  exactly  expressed  in  figures, 
for  the  number  3.141592  ....  by  which  d  is  multiplied,  is 
an  irrational  number,  and  is  commonly  denoted  by  -T  (the 
first  letter  of  -irepufttpeia,  meaning  circumference). 

Thus  C=tr  •  d,  or  2irr,  or  ir=C/d.  TT  then  expresses  the 
ratio  of  the  circumference  to  the  diameter  of  the  circle. 

Archimedes  (212  B.C.)  found  the  value  of  it  to  be  such  that 
3H<T<3l8  *>y  finding  the  value  of  P96  and  p9&. 

Ptolemaeus  (150  A.D.)  calculated  TT  =3.  14166. 

At  the  end  of  the  sixteenth  century  Vieta  (1579  A.D.)  found  the 
value  of  IT  to  10  decimal  places,  and  Ludolf  van  Eulen  (1540-1610)  to 
20,  32,  and  35  places.  The  value  of  ir  has  since  been  carried  out  to 
more  than  700  decimal  places. 


266 


Second-  Year  Mathematics 


EXERCISES 

1.  The   circumference   of   a  circle   is   loocm.     Find   the 
radius. 

2.  The  radius  of  a  circle  is  10.     Find  the  radius  of  a 
circle  (i)  4,  (2)  9,  (3)  25,  (4)  3,  (5)  17,  times  as  large  in  area. 

PROPOSITION  XVIII 

279.  Theorem:  The  area  of  a  regular  inscribed  polygon 
is  equal  to  the  product  of  one-half  of  the  perimeter  and  the  per- 
pendicular from  the  center  to  the  side* 


Proof:  Area  of  A  O  ~B  = 
Area  of  B  O  C  = 


A  B  (Fig.  315). 
B  C. 


FIG.  315 


Therefore  area  of  A  B  C  D  .  .  .  .=$h  (A  B+B  C  +  C  D 
.  .  .  .)  or  area  of  A  B  C  D  .  .  .  .=\h  -  p. 

PROPOSITION  XIX 

280.  Theorem:    The  area  of  a  regular  circumscribed  polygon 
is  the  product  of  one-half  the  perimeter  and  the  radius. 

The  proof  is  similar  to  that  of  Proposition  XVIII,  that 
area  ABC....   =\r  •  p.     See  Fig.  316. 


*This  perpendicular  is  called  the  apothem. 


Regular  Polygons  in  and  about  a  Circle  267 

Area  of  the  Circle 

281.  The  irrational  number  TT  has  now  been   denned  as 
the  limit  approached  by  the  series  of  numbers, 

4.000000  2.828427   .... 

3.464121    ....  3.000000 


3.141662   ....  3.141557   .... 

They  approach  each  other,  and  approach  TT,  more  and 
more  closely,  the  value  of  TT  always  lying  between  any  two 
corresponding  numbers. 

In  preceding  chapters  (pp.  127-212)  other  irrational  num- 

AC        — 

bers,  1/74,  l/Tss,    1/27,  -~^  =  V  2  (p.  126),  have  been  met. 
j>  C 

Their  value  also  was  approximated  by  rational  numbers. 
But  each  is,  as  in  the  case  of  TT,  the  limit  approached  by  its 
successive  approximations. 

Theorems  like  "The  area  of  a  rectangle  equals  the  product 
of  the  base  by  the  altitude"  (§  231),  can  now  be  more  clearly 
seen  to  be  true,  even  if  the  dimensions  be  irrational  numbers. 

The  dimensions  can  be  expressed  by  rational  numbers 
only  to  as  close  a  degree  of  approximation  as  may  be  wished, 
but  they  are  expressed  exactly  by  IT,  V  2,  1/12,  etc.,  and  the 
theorem  concerned  is  absolutely  true  when  they  are  thus  used. 
•The  logical  proof  of  this,  however,  would  involve  a  complete 
study  of  limits,  which  is  considered  beyond  the  province  of 
secondary-school  work,  as  is  plain  from  §  278. 

The  area  of  a  circle  lies  between  that  of  the  circumscribed 
polygon  and  that  of  the  inscribed  polygon.  Why?  As  the 
number  of  sides  is  increased  5  and  s  approach  the  area  of  the 
circle  and  h  equals  or  approaches  r,  so  that  S  and  5.  approach 
\r  •  c  to  as  great  a  degree  of  accuracy  as  may  be  desired;  or 
S  and  s  approach  \r  •  c  as  a  limit. 

This  leads  to  the  following: 


268  Second-Year  Mathematics 

PROPOSITION  XIX 

282.  Theorem:    The  area  of  a  circle  is  one-half  the  product 
of  the  circumference  and  the  radius,  i.  e.,  area  of  circle  =\c  •  r. 

•n-d2 

Show  that  the  area  of  a  circle  =irr2,  or . 

4 

EXERCISES 

1.  Prove  that  the  circumferences  of  two  circles  are  to  each 
other  as  the  radii,  or  as  the  diameters, 

2.  Prove  that  the  areas  of  two  circles  are  to  each  other  as 
the  squares  of  the  radii,  or  as  the  squares  of  the  diameters. 

3.  What  is  the  ratio  of  the  areas  of  two  circles  whose  radii 
are  5  in.  and  10  in.  ? 

4.  The  area  of  a  circle  is  64.     Find  the  diameter  and 
circumference. 

5.  What  is  the  area  of  the  ring  formed  by  two  concentric 
circles  whose  radii  are  5  in.  and  6  in.,  respectively  ?   a  in.  and 
&  in.,  respectively  ? 

6.  The  areas  of  two  circles  are  in  the  ratio  of  2  to  4.     What 
is  the  ratio  of  the  diameters  ? 

7.  The  circumference  of  a  circle  is  50  in.     What  is  the 
area? 

8.  The  diameter  of  a  circular  table  is  3  ft.     What  is  the 
circumference  of  the  table  ?   the  area  ? 

9.  The  radii  of  two  circles  are  to  each  other  as  3  :  5,  and 
their  combined  area  is  3850.     Find  the  radii  of  the  circles. 

TT  22 

Use   ir  =  —  . 

7 

10.  The  radii  of  two  circles  are  to  each  other  as  7  :  24, 
and  the  radius  of  a  circle  equivalent  to  their  sum  is  50.     Find 
the  radii  of  the  other  two  circles. 


Regular  Polygons  in  and  about  a  Circle  269 

283.  Problem:    To  find  the  area  of  a  sector  of  a  circle. 

Suggestion:  The  area  of  AA  O  C  =  iA~C  •  O  B  (Fig.  317).  If 
AC  isjbisected  at  D,  the  sum.  of  the  areas  of  the  two  AsAOD  +  DOC 
is}(Al)  +  DC)  •  OB'.  Why? 


FIG.  317 

If  the  number  of  divisions  of  A  C  be  indefinitely  increased, 
and  lines  be  drawn  to  the  points  of  division,  the  sum  of  the 
bases  of  the  As  will  approach  AC  as  a  limit  (?),  and  the 
common  altitude  of  the  As  will  approach  the  radius  as  a 
limit  (?).  Hence  the  sum  of  the  As  will  approach  the  area 
of  the  sector  as  a  limit,  and  will  approach  £AC  •  r  (?). 

Hence  the  area  of  the  sector  A  O  C=^A  C  •  r. 

This  may  be  stated  as  a  theorem: 

284.  The  area  of  a  sector  of  a  circle  is  equal  to  one-half  the 
product  of  the  arc  of  the  sector,  and  the  radius  of  the  circle. 

EXERCISES 

1.  The  radius  of  a  circle  is  100  ft.     The  length  of  the  arc 
of  a  sector  is  25  ft.     Find  the  area  of  the  sector. 

2.  The  radius  of  a  sector  is  9  in.,  its  area  is  72  sq.  in. 
Find  the  length  of  the  arc. 

3.  The  area  of  a  sector  is  96  sq.  ft.,  and  the  radius  is 
1 2  ft.     How  long  is  the  arc  ? 

4.  The  area  of  a  sector  is  a  sq.  ft.,  and  the  radius  is  r  ft. 
Find  the  length  of  the  arc. 


270  Second-Year  Mathematics 

5.  The  radius  of  a  circle  is  8  in.     Find  the  area  of  a  sector 
with  arc  36°. 

6.  Find  the  area  of  the  segment  whose  arc  is  36°,  in  a 
circle  of  radius  12  in. 

Notice  that  36  is  ^  of  the  circumference.     Thus  the  base  of  the 
triangle  is  the  side  of  a  regular  lo-side. 

7.  Find  the  area  of  a  segment  of  arc  72°,  in  a  circle  of 
radius  20. 

8.  The  area  of  a  circle  is  15,400  sq.  in.     Find  the  area 
of  a  segment  whose  arc  is  60°. 


CHAPTER  VIII 

PROBLEMS    AND    EXERCISES    IN    GRAPHIC  AND 
GEOMETRIC    ALGEBRA 

285.  In  the  preceding  chapters  algebraical  processes  have 
been  frequently  used  in  the  solution  of  geometrical  problems. 
This  has  been  done  by  working  with  the  numbers  representing 
the  lengths  of  line-segments,  the  areas  of  surfaces,  and  other 
magnitudes  in  the  geometrical  problem. 

On  the  other  hand,  geometrical  processes  are  of  value  in 
the  treatment  of  algebraical  problems,  as  has  been  shown, 
for  instance,  in  the  graphical  solution  of  equations. 

In  a  further  application  of  geometry  to  algebra,  the  different 
number  expressions  occurring  in  algebraical  problems  may  be 
represented  by  straight  line-segments.  To  enable  the  student 
to  do  this  will  be  one  of  the  purposes  of  this  chapter. 

286.  Graphical  problems  and  exercises. 

1.  The  lengths  of  the  sides  of  a  triangular  piece  of  land 
are  125  rd.,  54  rd.,  and  112  rd.     A  drawing  is  made  of  it  on 
a  sheet  of  paper,  the  longest  side  of  which  is  a  little  over  3  ft. 
How  long  will  the  sides  of  the  triangle  in  the  drawing  be  ? 

2.  The  dimensions  of  a  rectangle  are  12  and  17.     Without 
computing  the  length  of  the  shorter  side,  construct  a  rectangle 
similar  to  the  given  one  and  having  a  length  8  for  its  longer 
side. 

3.  Construct  a  rectangle  as  in  Problem  2  having  a  length 
8  for  its  shorter  side. 

4.  The  sides  of  a  triangle  are  14  in.,  10  in.,  and  19  in., 
respectively,  and  are  to  be  reduced  to  the  ratio  1:7.     Make 
an  accurate  construction  with  the  use  of  a  ruler.     In  what 
ratio  will  the  area  of  the  triangle  then  be  reduced  ? 

271 


272  Second-Year  Mathematics 

5.  Of  triangle  A  B  C  we  have  given  the  sides  A  B,  C  B, 
and  the  segment  A  D  that  the  bisector  of  angle  B  cuts  off 
on  the  side  A  C.     Construct  the  triangle. 

6.  Construct  a  triangle  similar  to  a  given  triangle,  and 
having  a  given  line  /  as  one  of  its  sides.     How  many  different 
triangles  are  possible,  if  the  given  triangle  is  scalene  ?    If  it  is 
isosceles  ?    If  it  is  equilateral  ? 

7.  The  lines  a,  b,  and  c  being  given,  construct  a  line  I  such 


a 

287.  To  construct  line-segments  of  lengths  of  fractions  of 
an  inch  not  marked  on  the  ruler. 

1.  To  construct  accurately  a  line  of  ^  of  an  inch. 
Calling  the  required  line  x,  we  have  x=%,  or  J=  -. 

OC 

Hence,  we  see  that  the  required  line  is  obtained  by  constructing  a 
fourth  proportional  to  3  lines  of,  respectively,  7  inches,  4  inches,  and 
i  inch. 

2.  Construct  the  lines  whose  lengths  are  given  by  the 
following  expressions,  the  letters  denoting  given  lines  and  the 
unit  being  chosen  arbitrarily: 

«^          (3)*  <5)f  (7)¥          (9)? 

S2J 

(\  ^p          /  N  5  i f\   7  /o\    <;          /    \  7^ 

2)  —          (4)  —          (6)  -  (8)  A         (I0)  —  • 

288.  The  graphic  representation  of  statistics. 

i.  Make  an  accurate  graph  of  the  following  population- 
data  for  Chicago,  representing  a  population  of  one  million  by 
a  line  of  3  inches: 

1872     .35  million      1882     .55  million     1892     1.45  million 
1874     -4        "  1884     .65       "          1894     1.55       " 

1876     .4        "  1886     .7  1896     1.6 

1878     .45      "  1888     .8        "          1898     1.85       " 

1880     .5        "  1890  1.2         "          1900     2.  " 


Problems  in  Graphic  and  Geometric  Algebra          273 


289.  Constructions. 

1.  To  construct  a  rectangle  whose  area  shall  be  24  sq.  in., 
and  whose  base  shall  be  7  in. 

2.  To  construct  a  rectangle  whose  altitude  is  given  and 
whose  area  is  equal  to  that  of  a  given  square  of  known  sides. 

3.  The  area  of  a  triangle  is  15  sq.  in.     Construct  a  line 
equal  to  its  altitude  when  the  base  is  4  in. 

4.  For  a  triangle  with  an  area  of  17  sq.  in.  construct  lines 
equal  to  the  altitudes  for  bases  of  length  2,  3,  4,  5  ....  13  in., 
and  graph  the  altitude  as  a  function  of  the  base.     How  does 
the  altitude  change  as  the  base  increases?    As  it  decreases? 

290.  The  graphing  of  literal  equations. 

i.  Graph  a  line  whose  equation  is  ax+by=c,  where  a, 
b,  and  c  represent  given  lines. 

We  find  the  following  pairs  of  values  from  the  given 
equation: 


if  x=o,  then  y  =  - 


and  if  y—o,  then  x=- 


a 
c        ,  c 


(0 

(2) 


We  construct  lines  equal  to  -  and  -,  respectively,  and  plot 

&  0 

the  points  (i)  and  (2)  (Fig.  318). 


FIG.  318 

Make  graphs  of  the  following  equations: 

2.  2#+4;y  =  7  4. 

3.  3^-2^  =  5  5.   - 

6.  2X— ry=s 


274  Second-Year  Mathematics 

291.  To  find  points  coney clic  with  given  points. 

i.  Having  given  3  points,  A,  B,  and  C,  to  find  points  which 
will  be  concyclic*  with  A,  B,  and  C  (see  Fig.  319). 

Draw  the  line  AB  and  any  line  through  C  meeting  A  B,  as 
at  P.  Now,  if  we  determine  a  point  D  on  C  P,  such  that 
AP  •  PB  =  CP  •  PD,  the  point  D  will  be  concyclic  with 
A,  B,  andC.  Why? 

X 


/ 
C' 

FIG.  319 

For  the  construction,  draw  AC;  lay  off  a  line  P  E,  equal 
to  P  B,  on  P  X.  Then,  draw  E  F  ||  A  C,  and  lay  off  P  D 
equal  to  P  F  on  P  X.  The  point  D  is  the  required  point. f 
Why? 

2.  Having  given  3  points  P,  Q,  and  R,  determine  a  num- 
ber of  points  concyclic  with  those  points,  without  first  drawing 
the  circle. 

292.  To  represent  areas  and  volumes  by  line-segments. 

i .  Represent  by  a  line  the  area  of  a  rectangle  whose  dimen- 
sions are  p  and  q. 

Denoting  the  area  by  A,  we  have  A=p  •  q  or  ~  =  —  .     Why ? 

P     A 
From  this  the  method  of  construction  is  evident. 

Construct  lines  whose  lengths  are  represented  by  the 
following  expressions,  taking  a  convenient  unit  of  length  in 
each  problem. 

*  Lying  on  the  same  circle. 

t  Admitting  the  use  of  a  pair  of  triangles  to  draw  parallel  lines,  the 
construction  is  very  rapid. 


Problems  in  Graphic  and  Geometric  Algebra          275 

2.  ab  5.  pqr  8.  a*bc 

3-  3^  6.  ^  9.  §*»;y 

2#Z  4VX 

4.    -  7.  -  10.  *abc 

5  <* 

1 1 .  Represent  by  a  straight  line  the  volume  of  a  rectangular 
block  whose   dimensions  are  given  lines  a,  b,  and  c  inches, 
representing  i  cubic  in.  by  a  line  of  i  in. 

12.  Represent  by  a  straight  line  the  volume  of  a  cube 
whose  edge  is  equal  to  a  given  line  x. 

13.  Make  an  accurate  graph  of  the  areas  of  squares  whose 
sides  are  equal  to  £";    J";    $";    £";    i";    i£";    i$";    if"; 
and  2". 

14.  Make   an   accurate  graph   of   the  volumes  of  cubes 
whose  edges  are  equal  to  \"\    \"\    \";    £";    i";    ij";    \\"\ 
if";  and  2". 

15.  Compare  the  two  graphs  of  Problems  13  and  14.    Does 
the  volume  of  a  cube  increase  more  or  less  rapidly  than  the 
area  of  a  square,  when  the  sides  of  each  are  increased  simul- 
taneously ? 

16.  Given  two  triangles  ABC  and  A'B'C',  having  Z_A  = 
Z.A'.     Construct  lines  representing  their  areas  and  the  pro- 
ducts A  B  -  A  C  and  A'B'  •  A'C'. 

Then  verify  the  proposition  that  two  triangles  which  have 
one  angle  equal  are  to  each  other  as  the  products  of  the  sides 
including  that  angle. 

293.  Different  methods  for  constructing  a  square  equal  to  a 
given  rectangle. 

i.  Construct  a  square  equal  in  area  to  a  rectangle  whose 
dimensions  are  a  and  b. 

By  what  relation  are  the  side  x  of  the  square  and  the  lines  a  and  b 
connected  ? 


276 


Second-Year  Mathematics 


2.  Construct  a  square  equal  in   area  to   a  rectangle  of 
dimensions  5  and  7. 

3.  Construct  a  square  equal  in  area  to  a  triangle  of  area  21. 

J) 


FIG.  320 

4.  To  construct  a  line,  x,  a  mean  proportional  to  two  given 
lines  a  and  b,  we  may  proceed  in  one  of  the  three  following 
ways  (see  p.  98,  Problem  i): 

(a)  On  a  line  X  Y,  lay  off  line-segments  A  B  and  B  C 
equal,  respectively,  to  a-  and  b  (Fig.  320).  Describe  a  circle 
on  A  C  as  a  diameter,  and  erect  line  B  D  perpendicular  to 
AC  at  B,  cutting  the  circle  at  D.  The  line  B  D  is  the 
required  line.  Prove,  using  §  132,  p.  98. 


FIG.  321 

(b)  On  a  line  X  Y  lay  off  line-segments  A  B  and  A  C 
equal  to  a  and  b,  respectively  (Fig.  321).  Describe  a  circle 
on  A  B  as  a  diameter  and  erect  a  line  C  D  _l_  A  B  at  C  meet- 


Problems  in  Graphic  and  Geometric  Algebra          277 

ing   the  circle   at   D.     The  line   A  B    is   the   required  line. 
Prove,  using  Problem  7,  p.  99. 

(c)  On  a  line  X  Y  (Fig.  322)  lay  off  line-segments  A  B  and 
A  C  equal  to  a  and  b,  respectively.  Draw  any  circle  passing 
through  C  and  B,  such  as  circle  O.  Draw  a  tangent  from 
A  to  the  circle.  The  line  A  D  will  be  the  required  line. 
Prove,  using  Proposition  IV,  p.  150. 


FIG.  322 

5.  Under  what  conditions  would  method  (6)  or  (c)  have 
advantage  over  method  (a)  ? 

294.  Construct  lines  that  are  mean  proportionals  to  the 
following  pairs  of  lines: 

1.  1 2"  and  1 1"  5.     5"  and  13" 

2.  5"  and  15"  6.     9"  and    8" 

3.  7"  and  14"  7.     i"  and    6" 

4.  4"  and    9"  8.  10"  and    7" 
Use  in  .every  case  the  method  best  suited  to  the  problem. 


278  Second-Year  Mathematics 

9.  Using  an  arbitrary  unit  of  length,  construct  a  line  equal 
to  1/3. 

Representing  the  required  line  by  x,  we  have  ^  =  1/3,  or  x2=$; 
i.e.,  i  :  x  =  x  :  3,  from  which  the  method  of  construction  becomes 
evident. 

10.  Construct  lines  equal  to: 

(1)  1/7  (4)  l/8_  (7)  i/H 

(2)  i/6_  (5)  1/iS  (8)  I/a 

(3)  1/12  (6)  !/a&  (9)   !/9. 

11.  Two  lines  a  and  &  being  given,  construct  lines  equal  to 
ab,  I/a,  1/6,  and  I/  06. 

12.  Prove  graphically  the  relation:    l/a&  =  l/a  •  1/6,  i.e., 
I/a  :  l/a6  =  i  :  1/6. 

13.  Three  lines  a,  6,  and  c  being  given,  construct  a  line  x 

equal  in  length  to:   •*/  —  . 

By    the   methods  of    §292    and    §293,  we   construct    a  line  y  = 

b* 

—  ,  then  construct  a  line  x  =  yay. 

14.  Construct  lines  represented  by  the  following  expressions, 
in  which  the  letters  and  numbers  represent  lines  that  are  given 
in  some  arbitrary  unit  of  length: 

(i)  1/|  (4)  2fl±l/*a  (7) 

(5) 


15.  Having  given  the  three  sides  of  a  triangle,  construct  a 
line  representing  its  area. 
295.  Miscellaneous  constructions. 

Construct  lines  equal  to  the  roots  of  the  following  equa- 
tions: 

1.  4#2  —  7=0  4.  ax2—  c=o 

2.  3#2=a  5.  4#2—  33;  —  1=0 

3.  3#2  —  2^  —  5=0  6.  5 


Problems  in  Graphic  and  Geometric  Algebra          279 

7.  The  hypotenuse  of  a  right  triangle  is  7;   the  sum  of  the 
other  two  sides  is  9.     Construct  the  triangle. 

8.  Given  a  triangle  ABC  whose  sides  are  4,  7,  and  8,  it 
is  required  to  construct  a  right  triangle  by  decreasing  each  of 
the  sides  of  triangle  A  B  C  by  some  fixed  amount. 

First  compute  the  lengths  of  the  sides  of  the  new  triangle,  then 
construct  it. 

9.  Construct  a   line,    x,    equal    to    —  ^L+^L\/^.    Then 
prove,  both  geometrically  and  algebraically,  that  the  line  x  is 
the  greater  part  of  the  line  L  divided  in  extreme  and  mean 
ratio. 

10.  Construct  a  circle  which  shall  pass  through  two  given 
points  and  be  tangent  to  a  given  line,  not  parallel  to  the  line 
joining    the    two    given    points.        (Make    use    of    §  293, 
Problem  4  (c). 

11.  In  the  right  triangle  ABC  (Fig.  323)  the  altitude  line 
AD  is  drawn;    then  D  D'     is  drawn   J_AC;    D'D"xCD. 
Express  the  lines  A  D,  D  D',  D'D",  etc.,  in  terms  of  a,  b, 
and  c.     Similarly,  the  lines  D  E',  E'E",  etc. 


12.  Three  lines,  p,  q,  and  r  being  given  (of  which  r  is  the 
greatest),  construct  lines  equal  to  ^—^  ,  «-  ,  *— ^  ,  etc 

13.  The  sides  of  an  isosceles  triangle  being  given,  it  is 
required  to  construct  lines  equal  to  the  radius  of  the  circum- 


280  Second-Year  Mathematics 

scribed  and  inscribed  circles,  without  constructing  the  triangle 
itself. 

296.  Construction  of   more  complicated  irrational   literal 
expressions. 

1.  Two  sides  of  a  right  triangle  are  a  and  b.     Construct 
the  hypotenuse,  x,  and  express  the  relation  which  exists  be- 
tween the  lines  x,  a,  and  b. 

2.  The  hypotenuse  and  one  side  of  a  right  triangle  being 
given  equal  to  c  and  a,  respectively,  construct  the  triangle  and 
express  the  relation,  which  exists  between  c,  a,  and  the  third 
side  b. 

3.  Construct  a  line  equal  to  (i)  ~\/p2+q2,  (2)  Vp2—  q2, 
p  and  q  being  given  lines.     What  relation  must  hold  between 
p  and  q  in  order  that  the  last  construction  may  be  possible  ? 


4.  Construct  a  line,  y,  equal  to  I/a2—  x2. 

Give  to  x  different  values,  and  graph  the  corresponding 
values  of  y. 

5.  Construct  a  line,  y,  equal  to  V/a2+x2. 

Give  to  x  different  values,  and  graph  the  corresponding 
values  of  y. 

6.  Using  the  constructions  of  §  293  to  §  296  combined,  con- 
struct a  line,  x,  equal  to  I/a2  —  2bc. 

Construct  a  line  y  equal  to  V  zbc;   then  x  =  V  a3—  y3. 

7.  Determine    lines    which   may   represent   the   following 
expressions  : 

(i)  I/a2  -25  (6)  x 


(2)  Vt*-rs  (7) 

(3)  Va*+bc  (8)  ia 

(4)  1/6"—  4ac 


(5)  p  +  1/pq  +  lab-pq 


Problems  in  Graphic  and  Geometric  Algebra         281 


8.  Construct  lines  equal  to  the  solutions  of  the  following 
equations: 

(i)  x2-4X  +  2=o  (6)  x2  +  (a-b)x+c=o 


(2)  #2  +  3#+a=o  (7)  x2—  tx+s=o 

(3)  T>x2—6x  +  2=o  (8)  ax2  +  2bx+c=o 

(4)  ax2+bx+c=o  (9)  x2-(a+b)x+(a-b)=o 

(5)  2X2+6x+3=o          (10)  2X2  +  (c-d)x  +  (c+d)=o. 
9.  Given  a  triangle  ABC.     To  determine  a  line  #,  such 

that  the  sides  of  the  triangle  ABC,  decreased  or  increased 
by  x,  may  form  the  sides  of  a  right  triangle. 


B 


10.  On  the  hypotenuse  of  the  right  triangle  ABC  (Fig.  324), 
a  segment  A  D  is  laid  off  equal  to  A  B.  D  E  is  drawn  _l_  A  C, 
and  A  E  is  drawn.  Then  A  D'  is  made  equal  to  A  E,  the  per- 
pendicular D'E'  is  erected  and  A  E'  drawn.  It  is  required  to 
express  A  E,  A  E',  etc.,  in  terms  of  the  sides  of  the  triangle. 


11.  Construct  a  line  equal  to  lx  aV ab. 

12.  Construct  a  line  equal  to  i/a. 

Notice  that  \f  a  =  V  i  •  V  i  •  a. 

297.  Construction  of  various  geometrical  figures. 

i.  Construct  a  triangle  similar  to  a  given  triangle  and 
having  an  area  equal  to  half  the  area  of  the  given  triangle; 
having  an  area  equal  to  J,  £,  of  the  given  triangle. 


282  Second-Year  Mathematics 

2.  Construct  a  square  equal  in  area  to  a  regular  triangle 
inscribed  in  a  circle  of  radius  2;  3;  4;  a. 

3.  Construct  an  equilateral  triangle  equal  in  area  to  a 
regular  hexagon  inscribed  in  a  circle  of  radius  2 ;   3;   5;  p. 

4.  A  regular  triangle  is  inscribed  in  a  circle  of  radius  3. 
It  is  required  to  construct  a  rectangle  having  base  5  and  equal 
in  area  to  the  triangle. 

5.  It  is  required  to  construct  a  circle  of  such  radius  that 
the  regular  triangle  inscribed  in  this  circle  shall  be  equal  in 
area  to  a  given  rectangle. 

6.  Construct  a  circle  of  such  radius  that  the  square  in- 
scribed in  this  circle  shall  be  equal  in  area  to  the  equilateral 
triangle  inscribed  in  a  given  circle. 

7.  Construct  a  circle  of  such  radius  that  the  area  of  the 
inscribed  regular  hexagon  shall  be  equal  to  the  area  of  a 
square  inscribed  in  a  given  circle. 

8.  The  hypotenuse  and  one  side  of  a  right  triangle  are 
given  lines.     Construct  a  square  equal  in  area  to  the  triangle. 

9.  Construct  a  circle  of   such  radius  that  the  inscribed 
square  shall  be  equal  in  area  to  the  area  of  a  given  triangle. 

Let  o=  altitude  and  5  =  base  of  the  given  triangle.     Construct  a 
circle  with  radius  =  Jy  ab.     Why  will  this  be  the  required  circle? 

10.  Construct  a  circle  of  such  radius  that  the  inscribed 
regular  triangle  shall  be  equal  in  area  to  a  given  triangle. 

11.  In  a  given  circle,  using  a  diameter  as  one  of  the  bases, 
inscribe  a  trapezoid  such  that  the  altitude  shall  be  one-half 
of  the  upper  base. 


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